case study questions on electric potential and capacitance class 12

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Class 12 Physics Chapter 2 Case Study Question Electrostatic Potential and Capacitance PDF Download

In Class 12 Boards there will be Case studies and Passage Based Questions will be asked, So practice these types of questions. Study Rate is always there to help you. Free PDF Download of CBSE Class 12 Physics Chapter 2 Electrostatic Potential and Capacitance Case Study and Passage Based Questions with Answers were Prepared Based on Latest Exam Pattern. Students can solve NCERT Class 12 Physics Electrostatic Potential and Capacitance  to know their preparation level.

In CBSE Class 12 Physics Paper, Students will have to answer some questions based on  Assertion and Reason . There will be a few questions based on case studies and passage-based as well. In that, a paragraph will be given, and then the MCQ questions based on it will be asked.

Electrostatic Potential and Capacitance Case Study Questions With answers

Here, we have provided case-based/passage-based questions for Class 12 Physics  Chapter 2 Electrostatic Potential and Capacitance

Case Study/Passage-Based Questions

Question 1:

Answer: (b) 20

(ii) A parallel plate capacitor with air between the plates has a capacitance of 8 pF. The separation between the plates is now reduced by half and the space between them is filled with a medium of dielectric constant 5. Calculate the value of capacitance of the capacitor in the second case.

Answer: (c) 80pF

(iii) A dielectric introduced between the plates of a parallel plate condenser

Answer: (d) increases the capacity of the condenser

(iv) A parallel plate capacitor of capacitance 1 pF has separation between the plates is d. When the distance of separation becomes 2d and wax of dielectric constant x is inserted in it the capacitance becomes 2 pF. What is the value of x?

Answer: (b) 4

Question 2:

When an insulator is placed in an external field, the dipoles become aligned. Induced surface charges on the insulator establish a polarization field Ē i  in its interior. The net field Ē in the insulator is the vector sum of Ē, and Ē i  as shown in the figure.

On the application of external electric field, the effect of aligning the electric dipoles in the insulator is called polarisation and the field Ē; is known as the polarisation field. The dipole moment per unit volume of the dielectric is known as  polarisation  ( P ). For linear isotropic dielectrics,  P =χE , where χ = electrical susceptibility of the dielectric medium.

(i) Which among the following is an example of polar molecule? (2) O₂ (b) H (c) N 2 (d) HCI

Answer: (d) HCI

(ii) When air is replaced by a dielectric medium of constant K, the maximum force of attraction between two charges separated by a distance (a) increases K times (b) remains unchanged (c) decreases K times (d) increases 2K times.

Answer: (c) decreases K times

(iii) Which of the following is a dielectric? (a) Copper (b) Glass (c) Antimony (Sb) (d) None of these

Answer: (c) Antimony (Sb)

(iv) For a polar molecule, which of the following statements is true ? (a) The centre of gravity of electrons and protons coincide. (b) The centre of gravity of electrons and protons do not coincide. (c) The charge distribution is always symmetrical. (d) The dipole moment is always zero.

Answer: (b) The centre of gravity of electrons and protons do not coincide.

(v) When a comb rubbed with dry hair attracts pieces of paper. This is because the (a) comb polarizes the piece of paper (b) comb induces a net dipole moment opposite to the direction of field (c) electric field due to the comb is uniform (d) comb induces a net dipole moment perpendicular to the direction of field

Answer: (a) comb polarizes the piece of paper

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12th Class Physics Electrostatics & Capacitance Question Bank

Done case based (mcqs) - electric potential and capacitance total questions - 50.

A) remain the same done clear

B) increase done clear

C) decrease done clear

D) become zero done clear

question_answer 2) Which of the following statement is not true?

A) Electrostatic force is a conservative force. done clear

B) Potential energy of charge q at a point is the work done per unit charge in bringing a charge from any point to infinity. done clear

C) Spring force and gravitational force are conservative force. done clear

D) Both [a] and [c] done clear

question_answer 3) Work done in moving a charge from one point to another inside a uniformly charged conducting sphere is

A) always zero done clear

B) non-zero done clear

C) may be zero done clear

D) none of these done clear

question_answer 4) The work done in bringing a unit positive charge from infinite distance to a point at distance x from a positive charge Q is W. Then the potential \[\phi \] at that point is

A) \[\frac{WQ}{x}\] done clear

B) W done clear

C) \[\frac{W}{x}\] done clear

D) WQ done clear

question_answer 5) If \[1\mu C\]charge is shifted from A to B and it is found that work done by an external force is \[40\,\,\mu J\]. In doing so against electrostatics force, the potential difference \[{{V}_{A}}-{{V}_{B}}\] is

A) 40V done clear

B) - 40V done clear

C) 20V done clear

D) - 60V done clear

A) \[-1\times {{10}^{8}}V\] done clear

B) \[1\times {{10}^{8}}V\] done clear

C) \[6\centerdot 4\times {{10}^{-19}}V\] done clear

D) \[-6\centerdot 4\times {{10}^{-19}}V\] done clear

question_answer 7) The Change in electric potential energy of the proton for displacement from A to B is

A) \[1\centerdot 6\times {{10}^{-11}}J\] done clear

B) \[0\centerdot 5\times {{10}^{-23}}J\] done clear

C) \[-1\centerdot 6\times {{10}^{-11}}J\] done clear

D) \[-3\centerdot 2\times {{10}^{-22}}J\] done clear

question_answer 8) The mutual electrostatic potential energy between two protons which are at a distance of \[9\times {{10}^{-15}}m\], in \[_{92}{{U}^{235}}\] nucleus is

A) \[1\centerdot 56\times {{10}^{-14}}J\] done clear

B) \[5\centerdot 5\times {{10}^{-14}}J\] done clear

C) \[2\centerdot 56\times {{10}^{-14}}J\] done clear

D) \[4\centerdot 56\times {{10}^{-14}}J\] done clear

question_answer 9) If a system consists of two charges \[4\,mC\] and \[-3\,mC\] with no external field placed at (-5 cm, 0, 0) and (5 cm, 0, 0) respectively. The amount of work required to separate the two charges infinitely away from each other is

A) \[-1\centerdot 1\,J\] done clear

B) 2J done clear

C) \[2\centerdot 5\,J\] done clear

D) 3 J done clear

question_answer 10) As the proton moves from P to Q, then

A) the potential energy of proton decreases done clear

B) the potential energy of proton increases done clear

C) the proton loses kinetic energy done clear

D) total energy of the proton increases. done clear

A) 1 done clear

B) 2 done clear

C) 3 done clear

D) Both [a] and [b] done clear

question_answer 12) The signs of charges \[{{Q}_{1}}\] and \[{{Q}_{2}}\] respectively are

A) positive and negative done clear

B) negative and positive done clear

C) positive and positive done clear

D) negative and negative done clear

question_answer 13) Which of the two charges \[{{Q}_{1}}\]and \[{{Q}_{2}}\] is greater in magnitude ?

A) \[{{Q}_{2}}\] done clear

B) \[{{Q}_{1}}\] done clear

C) Same done clear

D) Can't determined done clear

question_answer 14) Which of the following statement is not true?

A) Electrostatic force is a conservation force. done clear

B) Potential energy of charge q at a point is the work done per unit charge is bringing a charge from any point to infinity. done clear

C) When two like charges lie infinite distance apart, their potential energy is zero. done clear

D) Both [a] and [c]. done clear

question_answer 15) Positive and negative point charges of equal magnitude are kept at \[\left( 0,\,0,\,\frac{a}{2} \right)\] and \[\left( 0,\,0,\,\frac{-a}{2} \right)\]respectively. The work done by the electric field when another positive point charge is moved from (-a, 0, 0) to (0, a, 0) is

A) positive done clear

B) negative done clear

C) zero done clear

D) depends on the path connecting the initial and final positions. done clear

A) Equipotential surface due to a single point charge is spherical. done clear

B) Equipotential surface can be constructed for dipoles too. done clear

C) The electric field is normal to the equipotential surface through the point. done clear

D) The work done to move a test charge on the equipotential surface is positive. done clear

question_answer 17) Nature of equipotential surface for a point charge is

A) Ellipsoid with charge at foci done clear

B) Sphere with charge at the centre of the sphere done clear

C) Sphere with charge on the surface of the sphere done clear

D) Plane with charge on the surface. done clear

question_answer 18) A spherical equipotential surface is not possible

A) inside a uniformly charged sphere done clear

B) for a dipole done clear

C) inside a spherical condenser done clear

D) for a point charge done clear

question_answer 19) The work done in carrying a charge q once round a circle of radius a with a charge Q at its centre is

A) \[\frac{qQ}{4\pi {{\varepsilon }_{0}}a}\] done clear

B) \[\frac{qQ}{4\pi {{\varepsilon }_{0}}{{a}^{2}}}\] done clear

C) \[\frac{q}{4\pi {{\varepsilon }_{0}}a}\] done clear

D) zero done clear

question_answer 20) The work done to move a unit charge along an equipotential surface from P to Q

A) must be defined as \[-\int\limits_{P}^{Q}{\overrightarrow{E}\,}.\,d\overrightarrow{t}\] done clear

B) is zero done clear

C) can have a non-zero value done clear

D) both [a] and [b] are correct done clear

A) \[k{{e}^{2}}\] done clear

B) \[{{e}^{2}}/2\] done clear

C) \[-k{{e}^{2}}/2\] done clear

question_answer 22)   Four equal charges q each are placed at four comers of a square of side a each. Work done in carrying a charge -q from its centre to infinity is

A) zero done clear

B) \[\frac{\sqrt{2}q}{\pi {{\varepsilon }_{0}}a}\] done clear

C) \[\frac{\sqrt{2}q}{\pi {{\varepsilon }_{0}}a}\] done clear

D) \[\frac{{{q}^{2}}}{\pi {{\varepsilon }_{0}}a}\] done clear

question_answer 23) Two points A and B are located in diametrically opposite directions of a point charge of \[+2\,\,\mu C\]at distances 2 m and 1 m respectively from it. The potential difference between A and B is

A) \[3\times {{10}^{3}}V\] done clear

B) \[6\times {{10}^{4}}V\] done clear

C) \[-9\times {{10}^{3}}V\] done clear

D) \[-3\times {{10}^{3}}V\] done clear

question_answer 24) The point charges A = +3 nC and B = +1 nC are placed 5 cm apart in air. The work done to move charge B towards A by 1 cm is

A) \[2\centerdot 0\times {{10}^{-7}}\,J\] done clear

B) \[1\centerdot 35\times {{10}^{-7}}\,J\] done clear

C) \[2\centerdot 7\times {{10}^{-7}}\,J\] done clear

D) \[1\centerdot 21\times {{10}^{-7}}\,J\] done clear

question_answer 25) A charge Q is placed at the origin. The electric potential due to this charge at a given point in space is V. The work done by an external force in bringing another charge q from infinity up to the point is

A) \[\frac{V}{q}\] done clear

B) Vq done clear

C) \[V+q\] done clear

D) V done clear

A) 90 cm done clear

B) 45 cm done clear

C) 45 m done clear

D) 90m done clear

question_answer 27) How much charge should be placed on a capacitance of 25 pF to raise its potential to \[{{10}^{5}}V\]?

A) \[1\mu C\] done clear

B) \[1\centerdot 5\mu C\] done clear

C) \[2\,\,\mu C\] done clear

D) \[2\centerdot 5\mu C\] done clear

question_answer 28) Dimensions of capacitance is

A) \[\left[ M{{L}^{-2}}\,{{T}^{4}}{{A}^{2}} \right]\] done clear

B) \[\left[ {{M}^{-1}}\,{{L}^{-1}}\,{{T}^{3}}\,{{A}^{1}} \right]\] done clear

C) \[\left[ {{M}^{-1}}\,{{L}^{-2\,}}\,{{T}^{4}}\,{{A}^{2}} \right]\] done clear

D) \[\left[ {{M}^{0}}\,{{L}^{-2}}\,{{T}^{4}}\,{{A}^{1}} \right]\] done clear

question_answer 29)   Metallic sphere of radius R is charged to potential V. Then charge q is proportional to

A) V done clear

B) R done clear

C) both V and R done clear

D) none of these. done clear

question_answer 30) If 64 identical spheres of charge q and capacitance C each are combined to form a large sphere. The charge and capacitance of the sphere is

A) 64 q, C done clear

B) 16q, 4C done clear

C) 64q, 4C done clear

D) 16q, 64C done clear

A) Charge remains constant, potential decreases & capacitance increases done clear

B) Charge remains constant, potential increases & Capacitance decreases done clear

C) Charge increases, potential increases & Capacitance decreases done clear

D) Charge decreases, potential decreases & Capacitance increases. done clear

question_answer 32) In a parallel plate capacitor, the capacity increases if

A) area of the plate decreases done clear

B) distance between the plates increases done clear

C) area of the plate increases done clear

D) dielectric constant decreases. done clear

question_answer 33) A parallel plate capacitor has two square plates with equal and opposite charges. The surface charge densities on the plates are \[+\sigma \] and \[-\,\ sigma \]respectively. In the region between the plates the magnitude of the electric field is

A) \[\frac{\sigma }{2{{\varepsilon }_{0}}}\] done clear

B) \[\frac{\sigma }{{{\varepsilon }_{0}}}\] done clear

C) 0 done clear

question_answer 34) If a parallel plate air capacitor consists of two circular plates of diameter 8 cm. At what distance should the plates be held so as to have the same capacitance as that of sphere of diameter 20 cm ?

A) 9 mm done clear

B) 4 mm done clear

C) 8 mm done clear

D) 2 mm done clear

question_answer 35) If a charge of \[+2\centerdot 0\times {{10}^{-8}}\] is placed on the positive plate and a charge of \[-1\centerdot 0\times {{10}^{-8}}C\] on the negative plate of a parallel plate capacitor of capacitance \[1\centerdot 2\times {{10}^{-3}}\mu F\], then the potential difference developed between the plates is

A) \[6\centerdot 25V\] done clear

B) \[3\centerdot 0V\] done clear

C) \[12\centerdot 5V\] done clear

D) 25V done clear

A) 10 done clear

B) 20 done clear

C) 50 done clear

D) 100 done clear

question_answer 37) A parallel plate capacitor with air between the plates has a capacitance of 8 pF. The separation between the plates is now reduced half and the space between them is filled with a medium of dielectric constant 5. Calculate the value of capacitance of the capacitor in second case.

A) 8 pF done clear

B) 10 pF done clear

C) 80 pF done clear

D) 100 pF done clear

question_answer 38) A dielectric introduced between the plates of a parallel plate condenser

A) decreases the electric field between the plates done clear

B) increases the capacity of the condenser done clear

C) increases the charge stored in the condenser done clear

D) increases the capacity of the condenser done clear

question_answer 39) A parallel plate capacitor 1 pF has separation between the plates is d. When the distance of separation becomes 2d and wax of dielectric constant x is inserted in it, the capacitance becomes 2 pF. What is the value of x ?

A) 2 done clear

B) 4 done clear

C) 6 done clear

D) 8 done clear

question_answer 40) A parallel plate capacitor having area A and separated by distance a is filled by copper plate of thickness b. The new capacity is

A) \[\frac{{{\varepsilon }_{0}}A}{d+\frac{b}{2}}\] done clear

B) \[\frac{{{\varepsilon }_{0}}A}{2d}\] done clear

C) \[\frac{{{\varepsilon }_{0}}A}{d-b}\] done clear

D) \[\frac{2{{\varepsilon }_{0}}A}{d+\frac{b}{2}}\] done clear

A) \[3\mu J\] done clear

B) \[24\,\mu J\] done clear

C) \[30\,\mu J\] done clear

D) \[108\,\mu J\] done clear

question_answer 42) A capacitor of capacitance of \[10\,\mu F\] is charged to 10 V. The energy stored in it is

A) \[100\,\,\mu J\] done clear

B) \[500\,\mu J\] done clear

C) \[1000\,\mu J\] done clear

D) \[1\,\mu J\] done clear

question_answer 43) A parallel plate air capacitor has capacity C farad, potential V volt and energy E joule. When the gap between the plates is completely filled with dielectric

A) both V and E increase done clear

B) both V and E decrease done clear

C) V decreases, E increases done clear

D) V increases, E decreases done clear

question_answer 44) A capacitor with capacitance \[5\,\mu F\] is charged to \[5\,\mu C\]. If the plates are pulled apart to reduce the capacitance to \[2\mu F\], how much work is done?

A) \[6\centerdot 25\times {{10}^{-6}}J\] done clear

B) \[3\centerdot 75\times {{10}^{-6}}J\] done clear

C) \[2\centerdot 16\times {{10}^{-6}}J\] done clear

D) \[2\centerdot 55\times {{10}^{-6}}J\] done clear

question_answer 45) A metallic sphere of radius 18 cm has been given a charge of \[5\times {{10}^{-6}}\,C\]. The energy of the charged conductor is

A) \[0\centerdot 2J\] done clear

B) \[0\centerdot 6J\] done clear

C) \[1\centerdot 2J\] done clear

D) \[2\centerdot 4J\] done clear

A) \[{{O}_{2}}\] done clear

B) \[{{H}_{2}}\] done clear

C) \[{{N}_{2}}\] done clear

D) \[HCl\] done clear

question_answer 47) When air is replaced by a dielectric medium of constant K, the maximum force of attraction between two charges separated by a distance

A) increases K times done clear

B) remains unchanged done clear

C) decreases K times done clear

D) increases 2K times. done clear

question_answer 48) Which of the following is a dielectric?

A) Copper done clear

B) Glass done clear

C) Antimony (Sb) done clear

D) None of these done clear

question_answer 49) For a polar molecule, which of the following statements is true?

A) The centre of gravity' of electrons and protons Coincide done clear

B) The centre of gravity of electrons and protons do not coincide done clear

C) The charge distribution is always symmetrical done clear

D) The dipole moment is always zero done clear

question_answer 50) When a comb rubbed with dry hair attracts pieces of paper. This is because the

A) comb polarizes the piece of paper done clear

B) comb induces a net dipole moment opposite to the direction of field done clear

C) electric field due to the comb is uniform done clear

D) comb induces a net dipole moment perpendicular to the direction of field. done clear

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• Electrostatic Potential and Capacitance Important Questions for CBSE Class 12 Physics Chapter 2

Important Practice Problems for CBSE Class 12 Physics Chapter 2: Free PDF Download

Students of class 12 can find the important questions of Chapter 2 physics class 12 provided in a PDF format here. Scoring good marks in class 12 board exams is extremely important for every student. Hence, the important questions for class 12 physics chapter 2 - Electrostatic Potential and capacitance is made available to the students so that they can make a quick revision of all the vital topics and the important questions of chapter 2 physics class 12 and confidently ace the board exams. The physics class 12 chapter 2 important questions PDF can be easily downloaded from vedantu.com and students can refer to them anytime.

Download CBSE Class 12 Physics Important Questions 2024-25 PDF

Also, check CBSE Class 12 Physics Important Questions for other chapters:

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Study Important Questions for Class 12 Physics Chapter 2 - Electrostatic Potential and Capacitance

Very Short Answer Question (1 Mark Questions)

1. Why does the electric field inside a dielectric decrease when it is placed in an external electric field?

Ans: The electric field that is present inside a dielectric decreases when it is placed in an external electric field because of polarisation as it creates an internal electric field which is opposite to the external electric field inside a dielectric due to which the net electric field gets reduced.

2. What is the work done in moving a $2\mu C$ point change from corner A to corner B of a square ABCD when a $10\mu C$ charge exists at the centre of the square?

Ans: Point A & B are at the same distance from point O.

\[{{V}_{A}}={{V}_{B}}\]

$\Rightarrow \text{ Work done }=\text{0}$

3. Force of attraction between two point electric charges placed at a distance \[d\] in a medium is \[F\]. What distance apart should these be kept in the same medium, so that force between them becomes\[\frac{F}{3}\]?

Ans: If two point charges are \[{{q}_{1}}\] and \[{{q}_{2}}\] separated by distance \[d\], it can be expressed as:

\[F=\frac{K{{q}_{1}}{{q}_{2}}}{{{d}^{2}}}\]

Suppose if force becomes \[\frac{F}{3}\] let the distance be \[x\]

\[\frac{F}{3}=\frac{K{{q}_{1}}{{q}_{2}}}{{{x}^{2}}}\]

\[\Rightarrow \frac{K{{q}_{1}}{{q}_{2}}}{3{{d}^{2}}}=\frac{K{{q}_{1}}{{q}_{2}}}{{{x}^{2}}}\]

\[\Rightarrow {{x}^{2}}=3{{d}^{2}}\]

\[\Rightarrow x=\sqrt{3}d\]

4. The distance of the field point on the equatorial plane of a small electric dipole is halved. By what factors will the electric field due to the dipole changes?

Ans: The formula \[E\propto \frac{1}{{{r}^{3}}}\] gives the relation between electric field and distance. 

\[\therefore E\propto \frac{1}{{{\left( \frac{r}{2} \right)}^{3}}}\Rightarrow E\propto \frac{8}{{{r}^{3}}}\]

Therefore, the electric field becomes eight times larger.

5. The Plates of a charged capacitor are connected by a voltmeter. If the plates of the capacitor are moved further apart, what will be the effect on the reading of the voltmeter?

Ans: The relation between capacitance, area, distance and dielectric constant is \[C=\frac{A{{\in }_{0}}}{d}\Rightarrow C\propto \frac{1}{d}\]

Hence, if distance increases, capacitance decreases.

Since \[V=\frac{Q}{C}\] and charge on the capacitor is constant.

Hence reading of the voltmeter increases.

6. What happens to the capacitance of a capacitor when a dielectric slab is placed between its plates?

Ans: The introduction of dielectric in a capacitor will reduce the effective charge on plate and therefore will increase the capacitance.

Short Answer Questions (2 Marks Questions)

1. Show mathematically that the potential at a point on the equatorial line of an electric dipole is Zero?

Ans:  Electric potential at point on the equatorial line of an electric dipole can be expressed mathematically as:

\[V={{V}_{PA}}+{{V}_{PB}}\]

\[\Rightarrow V=\frac{K\left( -q \right)}{r}+\frac{K\left( +q \right)}{r}\]

\[\Rightarrow V=0\]

2. A parallel plate capacitor with air between the plates has a capacitance of \[8pF\]\[\left( 1pF={{10}^{-12}}F \right)\]. What will be the capacitance if the distance between the plates is reduced by half and the space between them is filled with a substance of dielectric constant \[6\]?

Ans: For air, capacitance can be expressed as, \[{{C}_{0}}=\frac{A{{\in }_{0}}}{d}\]

\[{{C}_{0}}=8\times pF=8\times {{10}^{-12}}F\]

Now \[{d}'=\frac{d}{2}\] and \[K=6\]

\[\Rightarrow {C}'=\frac{A{{C}_{0}}}{{{d}'}}\times 2\times K\]

\[\Rightarrow {C}'=8\times {{10}^{-12}}\times 2\times 6\]

\[\Rightarrow {C}'=96\times {{10}^{-12}}pF\]

3. Draw one equipotential surfaces (1) Due to uniform electric field (2) For a point charge \[\left( q<0 \right)\]?

Ans: The diagrams are given as:

4. If the amount of electric flux entering and leaving a closed surface are \[{{\phi }_{1}}\] and \[{{\phi }_{2}}\] respectively. What is the electric charge inside the surface?

Ans: Net flux can be given as \[={{\phi }_{2}}-{{\phi }_{1}}\]

Since \[\phi =\frac{q}{{{\in }_{0}}}\]

\[\Rightarrow Q=\left( {{\phi }_{2}}-{{\phi }_{1}} \right){{\in }_{0}}\]

The electric charge inside the surface can be given as:

\[\Rightarrow Q=\phi {{\in }_{0}}\]

5. A stream of electrons travelling with speed \[v\,m/s\] at right angles to a uniform electric field E is deflected in a circular path of radius \[r\]. Prove that \[\frac{e}{m}=\frac{{{v}^{2}}}{rE}\]?

Ans: The path of the electron that is travelling with velocity \[v\,m/s\] at right angle of \[\overline{E}\] is of circular shape.

It requires a centripetal in nature, \[F=\frac{m{{v}^{2}}}{r}\]

It is provided by an electrostatic force \[F=eE\]

\[\Rightarrow eE=\frac{m{{v}^{2}}}{r}\]

\[\Rightarrow \frac{e}{m}=\frac{{{v}^{2}}}{rE}\]

6. The distance between the plates of a parallel plate capacitor is \[d\]. A metal plate of thickness \[\left( \frac{d}{2} \right)\] is placed between the plates. What will be the effect on the capacitance?

Ans: For air \[{{C}_{0}}=\frac{A{{\in }_{0}}}{d}\]

Thickness \[t=\frac{d}{2}\] only when \[k=\infty \]

\[{{C}_{0}}=\frac{A{{\in }_{0}}}{d}\]

\[{{C}_{metal}}=\frac{A{{\in }_{0}}}{\left( d-t \right)}\]

\[\Rightarrow {{C}_{metal}}=\frac{A{{\in }_{0}}}{\left( d-\frac{d}{2} \right)}\]

\[\Rightarrow {{C}_{metal}}=2A{{\in }_{0}}\]

\[\Rightarrow {{C}_{metal}}=2{{C}_{0}}\]

Hence, capacitance will double.

7. Two charges \[2\mu C\] and \[-2\mu C\] are placed at points A and B \[6cm\] apart.

(a) Identify an equipotential surface of the system.

Ans: The situation is represented in the adjoining figure.

An equipotential surface is the plane on which the total potential is zero everywhere. This plane is normal to line AB. The plane is located at the mid-point of line AB because the magnitude of charges is the same.

(b) What is the direction of the electric field at every point on this surface?

Ans: The direction of the electric field at every point on this surface Is normal to the plane in the direction of AB.

8. In a Van de Graaff type generator a spherical metal shell is to be a 

\[15\times {{10}^{6}}V\]electrode. The dielectric strength of the gas surrounding the electrode is \[5\times {{10}^{7}}V{{m}^{-1}}\]. What is the minimum radius of the spherical shell required? (You will learn from this exercise why one cannot build an electrostatic generator using a very small shell which requires a small charge to acquire a high potential.)

Ans: Given that,

Potential difference is given as, \[V=15\times {{10}^{6}}V\]

Dielectric strength of the surrounding gas\[=5\times {{10}^{7}}V{{m}^{-1}}\]

Electric field intensity is given as, E = Dielectric strength = \[5\times {{10}^{7}}V{{m}^{-1}}\]

Minimum radius of the spherical shell required for the purpose is given by,

\[r=\frac{V}{E}\]

\[\Rightarrow r=\frac{15\times {{10}^{6}}}{5\times {{10}^{7}}}\]

\[\Rightarrow r=0.3m=30cm\]

Therefore, the minimum radius of the spherical shell required is \[30cm\].

9. A small sphere of radius \[{{r}_{1}}\] and charge \[{{q}_{1}}\] is enclosed by a spherical shell of radius \[{{r}_{2}}\] and charge \[{{q}_{2}}\]. Show that if \[{{q}_{1}}\] is positive, charge will necessarily flow from the sphere to the shell (when the two are connected by a wire) no matter what the charge \[{{q}_{2}}\] on the shell is.)

Ans: According to the statement of Gauss’s law, the electric field between a sphere and a shell is determined by the charge \[{{q}_{1}}\] on a small sphere. Therefore, the potential difference, \[V\], between the sphere and the shell is independent of charge \[{{q}_{2}}\]. For positive charge \[{{q}_{1}}\], potential difference \[V\]is always positive.

10. Describe schematically the equipotential surfaces corresponding to

(a) a constant electric field in the z-direction,

Ans: Equidistant planes which are parallel to the x-y plane are the equipotential surfaces.

(b) a field that uniformly increases in magnitude but remains in a constant (say, z) direction,

Ans: Planes parallel to the x-y plane are the equipotential surfaces with the exception that when the planes get closer, the field increases.

(c) a single positive charge at the origin, and

Ans: Concentric spheres centred at the origin are equipotential surfaces.

(d) a uniform grid consisting of long equally spaced parallel charged wires in a plane.

Ans: A periodically changing shape near the given grid is the equipotential surface. This shape gradually reaches the shape of planes parallel to the grid at a larger distance.

Short Answer Questions (3 Marks Questions)

1. Two dielectric slabs of dielectric constant \[{{K}_{1}}\] and \[{{K}_{2}}\] are filled in between the two plates, each of area \[A\], of the parallel plate capacitor as shown in the figure. Find the net capacitance of the capacitor? Area of each plate \[\frac{A}{2}\].

Ans: In the question, the two capacitors are in parallel

Net Capacitance, \[C={{C}_{1}}+{{C}_{2}}\] 

\[{{C}_{1}}=\frac{{{K}_{1}}{{\in }_{0}}\left( \frac{A}{2} \right)}{d}=\frac{{{K}_{1}}{{\in }_{0}}A}{2d}\]

\[{{C}_{2}}=\frac{{{K}_{2}}{{\in }_{0}}\left( \frac{A}{2} \right)}{d}=\frac{{{K}_{2}}{{\in }_{0}}A}{2d}\]

\[\Rightarrow C=\frac{{{K}_{1}}{{\in }_{0}}A}{2d}+\frac{{{K}_{2}}{{\in }_{0}}A}{2d}\]

\[\Rightarrow C=\frac{{{\in }_{0}}A}{2d}\left( {{K}_{1}}+{{K}_{2}} \right)\], which is the required capacitance.

2. Prove that the energy stored in a parallel plate capacitor is given by \[\frac{1}{2}C{{V}^{2}}\].

Ans: Let us suppose a capacitor is connected to a battery and it supplies a small amount of change \[dq\] at constant potential \[V\], then small amount of work done by the battery is given by

\[\Rightarrow dw=\frac{qc}{dq}\] …… (Since, \[q=CV\])

Total work done where the capacitor is fully changed to \[q\].

\[\int{dw=W}=\int_{0}^{q}{\frac{q}{c}dq}\]

\[\Rightarrow W=\int_{0}^{q}{qdq}\]

\[\Rightarrow W=\frac{{{q}^{2}}}{2C}=\frac{{{C}^{2}}{{V}^{2}}}{2C}\]

\[\Rightarrow W=\frac{1}{2}C{{V}^{2}}\]

This work done is stored in the capacitor in the form of electrostatic potential energy.

\[W=U=\frac{1}{2}C{{V}^{2}}\]

Hence proved.

3. State Gauss’s Theorem in electrostatics? Using this theorem, define an expression for the field intensity due to an infinite plane sheet of change of charge density \[\sigma C/{{m}^{2}}\].

Ans: Gauss’s Theorem has a statement that electric flux through a closed surface enclosing a charge \[q\] in vacuum is \[\frac{1}{{{\in }_{0}}}\] times the magnitude of the charge enclosed is \[\phi =\frac{q}{{{\in }_{0}}}\]

Let us consider a charge distributed over an infinite sheet of area S having surface change density\[\sigma C/{{m}^{2}}\]

To enclose the charge on sheet an imaginary Gaussian surface cylindrical in shape can be assumed and it is divided into three sections \[{{S}_{1}}\], \[{{S}_{2}}\] and \[{{S}_{3}}\].

According to Gauss’s theorem we can infer,

\[\phi =\frac{q}{{{\in }_{0}}}=\frac{\sigma S}{{{\in }_{0}}}\] …… (1)

We know that, \[\phi =\int{E\cdot ds}\]

For the given surface \[\phi =\int\limits_{{{S}_{1}}}{\overrightarrow{E}\cdot \overrightarrow{ds}}+\int\limits_{{{S}_{2}}}{\overrightarrow{E}\cdot \overrightarrow{ds}}+\int\limits_{{{S}_{3}}}{\overrightarrow{E}\cdot \overrightarrow{ds}}\]

\[\Rightarrow \phi =\int\limits_{{{S}_{1}}}{E\cdot ds}\cos {{0}^{\circ }}+\int\limits_{{{S}_{3}}}{E\cdot ds}\cos {{0}^{\circ }}\] …… \[\left( \because \theta ={{90}^{\circ }}\,for\,{{S}_{2}}\,so\,\int\limits_{{{S}_{2}}}{\overrightarrow{E}\cdot \overrightarrow{ds}}=0 \right)\]

\[\Rightarrow \phi =E\int\limits_{{{S}_{1}}}{ds}+E\int\limits_{{{S}_{3}}}{ds}\]

\[\Rightarrow \phi =E\left[ \int_{0}^{S}{ds}+\int_{0}^{S}{ds} \right]\]

\[\Rightarrow \phi =E\cdot 2S\] ……(2)

From both equation (1) & (2)

\[E=\frac{\sigma }{2{{\in }_{0}}}\]

\[\Rightarrow E\cdot 2S=\frac{\sigma S}{{{\in }_{0}}}\], which is the required field intensity.

4. Derive an expression for the total work done in rotating an electric dipole through an angle \[\theta \] in a uniform electric field? \[E=1.5\times {{10}^{-8}}J\]

Ans: We know \[\tau =PE\sin \theta \]

If an electric dipole is rotated through an angle \[\theta \] against the torque, then small amount of work done is

\[dw=\tau d\theta =PE\sin \theta d\theta \]

For rotating through an angle \[\theta \] from \[{{90}^{\circ }}\]

\[W=\int_{{{90}^{\circ }}}^{\theta }{{}}PE\sin \theta \]

5. If \[{{C}_{1}}=3pF\] and \[{{C}_{2}}=2pF\], calculate the equivalent capacitance of the given network between A and B?

Ans: Since \[{{C}_{1}}\], \[{{C}_{2}}\]  and \[{{C}_{1}}\] are in series

\[\frac{1}{C}=\frac{1}{{{C}_{1}}}+\frac{1}{{{C}_{2}}}+\frac{1}{{{C}_{3}}}\]

\[\Rightarrow \frac{1}{C}=\frac{1}{3}+\frac{1}{2}+\frac{1}{3}\]

\[\Rightarrow \frac{1}{C}=\frac{2+3+2}{6}\]

\[\Rightarrow C=\frac{6}{7}pF\]

\[{{C}_{2}}\] and C are in series

\[\Rightarrow {C}'=\frac{2}{1}+\frac{6}{7}=\frac{14+6}{7}=\frac{20}{7}pF\]

\[{{C}_{1}},{C}'\] and \[{{C}_{1}}\] are in series

\[\frac{1}{{{C}_{net}}}=\frac{1}{{{C}_{1}}}+\frac{1}{{{C}'}}+\frac{1}{{{C}_{1}}}\]

\[\Rightarrow \frac{1}{{{C}_{net}}}=\frac{1}{3}+\frac{7}{20}+\frac{1}{3}\]

\[\Rightarrow \frac{1}{{{C}_{net}}}=\frac{20+21+20}{60}=\frac{61}{60}\]

\[\Rightarrow \frac{1}{{{C}_{net}}}=\frac{61}{60}pF\Rightarrow {{C}_{net}}=\frac{60}{61}pF\], which is the net capacitance.

6. Prove that energy stored per unit volume in a capacitor is given by \[\frac{1}{2}{{\in }_{0}}{{E}^{2}}\], where E is the electric field of the capacitor.

Ans: We know capacitance of a parallel plate capacitor, \[C=\frac{A{{\in }_{0}}}{d}\], electric field in between the plates 

Energy stored per unit volume\[=\frac{Energy\,stored}{Volume}\]

Energy stored per unit volume\[=\frac{\frac{1}{2}\frac{{{q}^{2}}}{C}}{Ad}\] (Volume of the capacitor = Ad)

\[=\frac{\frac{1}{2}\times \frac{{{\left( \sigma A \right)}^{2}}}{\frac{A{{\in }_{0}}}{d}}}{Ad}=\frac{\frac{1}{2}\times {{E}^{2}}{{\in }^{2}}{{A}^{2}}d}{{{A}^{2}}{{\in }_{0}}d}\]

\[\Rightarrow \]Energy stored/volume\[=\frac{1}{2}{{\in }_{0}}{{E}^{2}}\]. 

Therefore, proved.

7. Keeping the voltage of the charging source constant. What would be the percentage change in the energy stored in a parallel plate capacitor if the separation between its plates were to be decreased by \[10%\]?

Ans: We have the relation, \[U=\frac{1}{2}C{{V}^{2}}\]

For parallel plate \[U=\frac{1}{2}\frac{A{{\in }_{0}}}{d}{{V}^{2}}\]

When \[{d}'=d-\frac{10}{100}\times d=0.9d\]

Then \[{U}'=\frac{1}{2}\frac{A{{\in }_{0}}}{0.9d}{{V}^{2}}\]

Change in energy\[={U}'-U=\frac{1}{2}\frac{A{{\in }_{0}}}{d}\left( \frac{1}{0.9}-1 \right)\]

\[={U}'-U=u\left( \frac{0.1}{0.9} \right)=\frac{U}{9}\]

\[%change=\frac{{U}'-U}{U}\times 100%=\frac{U}{9}\times \frac{1}{U}\times 100%=\frac{100}{9}%\]

\[\Rightarrow \% change=11.1%\]

8. Two identical plane metallic surfaces A and B are kept parallel to each other in air separated by a distance of \[1.0cm\] as shown in figure.

Surface A is given a positive potential of \[10V\] and the outer surface of B is earthed.

(a) What is the magnitude and direction of the uniform electric field between point Y and Z? What is the work done in moving a charge of \[20\mu C\] from point X to Y?

Ans:   Since \[E=\frac{dv}{dr}=\frac{10V}{1\times {{10}^{-2}}}=1000\frac{V}{M}\]

Since surface A is an equipotential surface i.e., \[\Delta V=0\]

$\therefore$ Work done in moving a charge of $20 \mu C$ from X to Y = Zero.

(b) Can we have non-zero electric potential in space, where electric field strength is zero?

Ans:  Since surface A is an equipotential surface i.e., \[\Delta V=0\]

$\therefore$ Work done from X to Y = Zero.

\[E=\frac{-dv}{dr}\] if \[E=0\]

\[\frac{dv}{dr}=0\Rightarrow dv=0\] or V=constant (non-zero)

So, we can have non-zero electric potential, where the electric field is zero.

9. A regular hexagon of side \[10cm\] has a charge \[5\mu C\] at each of its vertices. Calculate the potential at the centre of the hexagon.

Ans: The given figure shows six equal numbers of charges \[q\], at the vertices of a regular hexagon.

Charge, \[q=5\mu C=5\times {{10}^{-5}}C\]

Side of the hexagon, \[AB=BC=CD=DE=EF=FA=10cm\]

Distance of each vertex from center \[O\], \[d=10cm\]

Electric potential at point \[O\],

\[V=\frac{6\times q}{4\pi {{\in }_{0}}d}\]

\[{{\in }_{0}}\]permittivity of free space

\[\frac{1}{4\pi {{\in }_{0}}}=9\times {{10}^{9}}N{{C}^{-2}}{{m}^{-2}}\]

\[V=\frac{6\times 9\times {{10}^{9}}\times 5\times {{10}^{-5}}}{0.1}\]

\[\Rightarrow V=2.7\times {{10}^{6}}V\]

Therefore, the potential at the centre of the hexagon is \[2.7\times {{10}^{6}}V\].

10. A spherical conductor of radius \[12cm\] has a charge of \[1.6\times {{10}^{-7}}C\] distributed uniformly on its surface. What is the electric field? 

(a) Inside the sphere.

Ans: Radius of the spherical conductor is given as, \[r=12cm=0.12m\]

Charge is uniformly distributed over the conductor and can be given as, \[q=1.6\times {{10}^{-7}}C\]

Electric field inside a spherical conductor is zero. This is because if there is a field inside the conductor, then charges will move to neutralize it.

(b) Just outside the sphere  

Ans: Electric field E just outside the conductor is given by the relation,

\[E=\frac{q}{4\pi {{\in }_{0}}{{r}^{2}}}\]

\[{{\in }_{0}}=\]permittivity of free space

\[\Rightarrow E=\frac{1.6\times {{10}^{-7}}\times 9\times {{10}^{-9}}}{{{\left( 0.12 \right)}^{2}}}\]

\[\Rightarrow E={{10}^{5}}N{{C}^{-1}}\]

Therefore, the electric field just outside the sphere \[={{10}^{5}}N{{C}^{-1}}\]

(c) At a point \[18cm\] from the centre of the sphere?

Distance of the point from the centre, \[d=18cm=0.18m\]

\[E=\frac{q}{4\pi {{\in }_{0}}{{d}^{2}}}\]

 \[\Rightarrow E=\frac{9\times {{10}^{9}}\times 1.6\times {{10}^{-7}}}{\left( 18\times {{10}^{-2}} \right)}\]

 \[\Rightarrow E=4.4\times {{10}^{4}}N/C\]

Therefore, the electric field at a point 18 cm from the centre of the sphere is \[4.4\times {{10}^{4}}N/C\].

11. Three capacitors each of capacitance \[9pF\] are connected in series.

(a) What is the total capacitance of the combination

Ans: Capacitance of each of the three capacitors, \[C=9pF\]

Equivalent capacitance \[{C}'\] of the combination of the capacitors is given by the relation,

\[\frac{1}{C}=\frac{1}{{{C}'}}+\frac{1}{{{C}'}}+\frac{1}{{{C}'}}\]

\[\Rightarrow \frac{1}{C}=\frac{1}{9}+\frac{1}{9}+\frac{1}{9}=\frac{3}{9}\]

\[\Rightarrow C=3\mu F\]

Therefore, the total capacitance of the combination is \[3\mu F\].

(b) What is the potential difference across each capacitor if the combination is connected to a \[120V\] supply?

Ans:   Supply voltage is given as, \[V=100V\]

Potential difference \[V\]  across each capacitor is equal to one-third of the supply voltage.

\[{V}'=\frac{V}{3}=\frac{120}{3}=40V\]

Therefore, the potential difference across each capacitor is \[40V\].

12. Three capacitors of each capacitance \[2pF\] , \[3pF\] and \[4pF\]  are connected in parallel.

(a) What is the total capacitance of the combination?

Ans: Capacitance of the given capacitors are

\[{{C}_{1}}=2pF\]

\[{{C}_{2}}=3pF\]

\[{{C}_{3}}=4pF\]

For the parallel combination of the capacitors, equivalent capacitor \[{C}'\] is given by the algebraic sum,

\[{C}'=\left( 2+3+4 \right)pF\]

Therefore, the total capacitance of the combination is \[9pF\].

(b) Determine the charge on each capacitor if the combination is connected to a \[100V\] supply.

Ans:  Supply voltage, \[V=100V\]

The voltage through all the three capacitors is same as \[V=100V\] 

Charge on a capacitor of capacitance \[C\] and potential difference \[V\] is given by the relation,

\[q=CV\] …… (1)

For \[C=2pF\],

Charge\[=VC=100\times 2=200pC=2\times {{10}^{-10}}C\]

For \[C=3pF\], 

Charge\[=VC=100\times 3=300pC=3\times {{10}^{-10}}C\]

For \[C=4pF\], 

Charge\[=VC=100\times 4=400pC=4\times {{10}^{-10}}C\]

13. In a parallel plate capacitor with air between the plates, each plate has an area and the distance between the plates is \[3mm\]. Calculate the capacitance of the capacitor. If this capacitor is connected to a \[100V\] supply, what is the charge on each plate of the capacitor?

Ans: Area of each plate of the parallel plate capacitor, \[A=6\times {{10}^{-3}}{{m}^{2}}\]

Distance between the plates, \[d=3mm=3\times {{10}^{-3}}m\]

Supply voltage, \[V=100V\]

Capacitance \[C\] of a parallel plate capacitor is given by,

\[C=\frac{A{{\in }_{0}}}{d}\]

\[\Rightarrow {{\in }_{0}}=8.854\times {{10}^{-12}}{{N}^{-1}}{{m}^{-2}}{{C}^{-2}}\]

\[\Rightarrow C=\frac{8.854\times {{10}^{-12}}\times 6\times {{10}^{-3}}}{3\times {{10}^{-3}}}\]

\[\Rightarrow C=17.71\times {{10}^{-12}}F\Rightarrow C=17.71pF\]

Potential \[V\] is related with the charge \[q\] and capacitance \[C\] as

\[V=\frac{q}{C}\Rightarrow q=VC\Rightarrow q=1.771\times {{10}^{-9}}C\]

\[\Rightarrow q=100\times 17.71\times {{10}^{-12}}C=17.71pF\]

Therefore, the capacitance of the capacitor is \[17.71pF\] and charge on each plate is \[1.771\times {{10}^{-9}}\].

14. Explain what would happen if in the capacitor given in Exercise 2.8, a \[3mm\] thick mica sheet (of dielectric constant \[=6\]) were inserted between the plates,

(a) While the voltage supply remained connected.

Ans: Dielectric constant of the mica sheet is given in the question as, \[k=6\]

Initial capacitance is provided as, \[C=1.771\times {{10}^{-11}}\]

New capacitance is given as, \[{C}'=kC=6\times 1.771\times {{10}^{-11}}=106pF\]

Supply voltage is given as, \[V=100V\]

New charge \[{q}'={C}'V=6\times 1.7717\times {{10}^{-9}}=1.06\times {{10}^{-8}}Cs\]

Potential across the plates remains \[100V\].

(b). After the supply was disconnected.

Ans: Dielectric constant is given in the question as, \[k=6\]

Initial capacitance, \[C=1.771\times {{10}^{-11}}F\]

New capacitance is given as \[{C}'=kC=6\times 1.771\times {{10}^{-11}}F=106pF\]

If supply voltage is removed, then there will be no effect on the amount of charge in the plates.

Charge \[=1.771\times {{10}^{-9}}C\]

Potential across the plates is given by,

\[V=\frac{q}{{{C}'}}\]

\[\Rightarrow V=\frac{1.771\times {{10}^{-9}}}{106\times {{10}^{-12}}}\]

\[\Rightarrow V=16.7V\]

15. A \[12pF\] capacitor is connected to a \[50V\] battery. How much electrostatic energy is stored in the capacitor?

Ans: Capacitor of the capacitance is given as, \[C=12pF=12\times {{10}^{-12}}F\]

Potential difference is provided as, \[V=50V\]

Electrostatic energy stored in the capacitor is given by the relation,

\[{{E}_{B}}=\frac{{{q}_{2}}}{4\pi {{\in }_{0}}{{\left( BZ \right)}^{2}}}\]

\[\cos \theta =\frac{0.10}{0.18}=\frac{5}{9}=0.5556\]

\[\theta ={{\cos }^{-1}}0.5556={{56.25}^{\circ }}\]

\[\Rightarrow 2\theta ={{112.5}^{\circ }}\]

\[\Rightarrow \cos 2\theta =-0.38\]

\[\overline{{{E}_{2}}}-\overline{{{E}_{1}}}=\frac{\sigma }{2{{\in }_{0}}}\hat{n}+\frac{\sigma }{2{{\in }_{0}}}\hat{n}=\frac{\sigma }{{{\in }_{0}}}\hat{n}\]

\[\Rightarrow E\left( 2\pi dL \right)=\frac{\lambda L}{{{\in }_{0}}}\]

\[\Rightarrow E\left( 2\pi dL \right)=\frac{{{q}_{1}}{{q}_{2}}}{4\pi {{\in }_{0}}{{d}_{1}}}-27.2eV\]

\[\Rightarrow E\left( 2\pi dL \right)=\frac{9\times {{10}^{9}}\times {{\left( 1.6\times {{10}^{-19}} \right)}^{2}}}{1.06\times {{10}^{-10}}}-27.2eV\]

\[\Rightarrow E\left( 2\pi dL \right)=21.73\times {{10}^{-10}}J-27.2eV\]

\[\Rightarrow E\left( 2\pi dL \right)=13.58eV-27.2eV\]

\[\Rightarrow E\left( 2\pi dL \right)=13.58eV\]

\[E=\frac{1}{2}C{{V}^{2}}\]

\[\Rightarrow E=\frac{1}{2}\times 12\times {{10}^{-12}}\times {{\left( 50 \right)}^{2}}\]

\[\Rightarrow E=1.5\times {{10}^{-8}}J\]

Therefore, the electrostatic energy stored in the capacitor is \[1.5\times {{10}^{-8}}J\].

16. A \[600pF\]capacitor is charged by a \[200V\] supply. It is then disconnected from the supply and is connected to another uncharged \[600pF\] capacitor. How much electrostatic energy is lost in the process?

Capacitor of the capacitance, \[C=600pF\] 

Potential difference, \[V=200V\]

Electrostatic energy stored in the capacitor is given by,

\[\Rightarrow E=\frac{1}{2}\times \left( 600\times {{10}^{-12}} \right)\times {{\left( 200 \right)}^{2}}\]

\[\Rightarrow E=1.2\times {{10}^{-5}}J\]

If supply is disconnected from the capacitor and another capacitor of capacitance \[C=600pF\] is connected to it, then equivalent capacitance \[{C}'\] of the combination is given by,

\[\frac{1}{{{C}'}}=\frac{1}{C}+\frac{1}{C}\]

\[\Rightarrow \frac{1}{{{C}'}}=\frac{1}{600}+\frac{1}{600}=\frac{2}{600}=\frac{1}{300}\]

\[\Rightarrow {C}'=300pF\]

New electrostatic energy can be calculated as

\[{E}'=\frac{1}{2}{C}'\times {{V}^{2}}\]

\[{E}'=\frac{1}{2}\times 300\times {{\left( 200 \right)}^{2}}\]

\[{E}'=0.6\times {{10}^{-5}}J\]

Loss in electrostatic energy can be given as \[=E-{E}'\]

\[\Rightarrow E-{E}'=1.2\times {{10}^{-5}}-0.6\times {{10}^{-5}}\]

\[\Rightarrow E-{E}'=0.6\times {{10}^{-5}}\]

\[\Rightarrow E-{E}'=6\times {{10}^{-6}}J\]

Therefore, the electrostatic energy lost in the process is \[6\times {{10}^{-6}}J\].

17. A spherical conducting shell of inner radius \[{{r}_{1}}\] and outer radius \[{{r}_{2}}\] has a charge

(a) A charge \[q\] is placed at the centre of the shell. What is the surface charge density on the inner and outer surfaces of the shell?

Ans: Charge placed at the centre of a shell is \[+q\]. Hence a charge of magnitude \[-q\] will be induced to the inner surface of the shell. Therefore, total charge on the inner surface of the shell is \[-q\].

Surface charge density at the inner surface of the shell can be given by the relation,

\[{{\sigma }_{1}}=\frac{Total\,Charge}{Outer\,Surface\,Area}=\frac{-q}{4\pi {{r}_{1}}^{2}}\]

\[{{\sigma }_{2}}=\frac{Total\,Charge}{Outer\,Surface\,Area}=\frac{-q}{4\pi {{r}_{2}}^{2}}\]

A charge of \[+q\] is induced on the outer surface of the shell. A charge of magnitude \[Q\] is placed on the outer surface of the shell. Therefore, total charge on the outer surface of the shell is \[Q+q\]. Surface charge density at the outer surface of the shell.

(b) Is the electric field inside a cavity (with no charge) zero, even if the shell is not spherical, but any irregular shape? Explain.

Ans:   Yes, the electric field intensity which is inside a cavity is zero, even if the shell is not spherical and has any irregular shape. Now, take a closed loop such that a part of it is inside the cavity along a field line while the rest is inside the conductor. Net work done by the field in carrying a test charge over a closed loop will be zero due to the field inside the conductor being zero. Therefore, the electric field is zero, whatever the shape.

18. If one of the two electrons of a \[{{H}_{2}}\] molecule is removed, we get a hydrogen molecular ion \[{{H}_{2}}^{+}\]. In the ground state of an \[{{H}_{2}}^{+}\], the two protons are separated by roughly \[1.5{{A}^{\circ }}\], and the electron is roughly \[1{{A}^{\circ }}\] from each proton. Determine the potential energy of the system. Specify your choice of the zero of potential energy.

Ans: The system of two protons and one electron is represented in the given figure:

Charge on proton 1, \[{{q}_{1}}=1.6\times {{10}^{-19}}C\]

Charge on proton 2, \[{{q}_{2}}=1.6\times {{10}^{-19}}C\]

Charge on electron, \[{{q}_{3}}=-1.6\times {{10}^{-19}}C\]

Distance between protons 1 and 2, \[{{d}_{1}}=1.5\times {{10}^{-10}}m\]

Distance between proton 1 and electron, \[{{d}_{2}}=1\times {{10}^{-10}}m\]

Distance between protons 2 and electron, \[{{d}_{3}}=1\times {{10}^{-10}}m\]

The potential energy at infinity is zero.

Potential energy of the system,

\[V=\frac{{{q}_{1}}{{q}_{2}}}{4\pi {{\in }_{0}}{{d}_{1}}}+\frac{{{q}_{1}}{{q}_{3}}}{4\pi {{\in }_{0}}{{d}_{3}}}+\frac{{{q}_{3}}{{q}_{1}}}{4\pi {{\in }_{0}}{{d}_{2}}}\]

Substituting \[\frac{1}{4\pi {{\in }_{0}}d}=9\times {{10}^{9}}N{{m}^{2}}{{C}^{-2}}\]

\[V=\frac{9\times {{10}^{9}}\times {{10}^{-19}}}{{{10}^{-10}}}\left[ -{{\left( 16 \right)}^{2}}+\frac{{{\left( 1.6 \right)}^{2}}}{1.5}-{{16}^{2}} \right]\]

$\Rightarrow V=-30.7\times {{10}^{-19}}J $

$\Rightarrow V=-19.2eV$

Therefore, the potential energy of the system is \[-19.2eV\].

19. Two charged conducting spheres of radii \[a\] and \[b\] are connected to each other by a wire. What is the ratio of electric fields at the surfaces of the two spheres? Use the result obtained to explain why charge density on the sharp and pointed ends of a conductor is higher than on its flatter portions?

Ans: Let \[a\] be the radius of a sphere \[A\], \[{{Q}_{A}}\] be the charge on the sphere, and \[{{C}_{A}}\] be the capacitance of the sphere. Let \[b\] be the radius of a sphere \[B\], \[{{Q}_{B}}\] be the charge on the sphere, and \[{{C}_{B}}\] be the capacitance of the sphere. Since the two spheres are connected with a wire, their potential \[V\] will become equal.

Let \[{{E}_{A}}\] be the electric field of sphere \[A\] and \[{{E}_{B}}\] be the electric field of sphere \[B\]. Therefore, their ratio,

\[\frac{{{E}_{A}}}{{{E}_{B}}}=\frac{{{Q}_{A}}}{4\pi {{\in }_{0}}{{a}_{2}}}\times \frac{{{b}^{2}}4\pi {{\in }_{0}}}{{{Q}_{B}}}\]

\[\Rightarrow \frac{{{E}_{A}}}{{{E}_{B}}}=\frac{{{Q}_{A}}}{{{Q}_{B}}}\times \frac{{{b}^{2}}}{{{a}_{2}}}\]

However, 

\[\frac{{{Q}_{A}}}{{{Q}_{B}}}=\frac{{{C}_{A}}V}{{{C}_{B}}V}\]

\[\frac{{{C}_{A}}}{{{C}_{B}}}=\frac{a}{b}\]

\[\frac{{{Q}_{A}}}{{{Q}_{B}}}=\frac{a}{b}\]

Putting the value of (2) in (1), we obtain

\[\frac{{{E}_{A}}}{{{E}_{B}}}=\frac{a}{b}\frac{{{b}^{2}}}{{{a}^{2}}}=\frac{b}{a}\]

Therefore, the ratio of the electric field at the surface is \[\frac{b}{a}\].

20. What is the area of the plates of a \[2F\] parallel plate capacitor, given that the separation between the plates is \[0.5cm\]? (You will realize from your answer why ordinary capacitors are in the range \[\mu F\] or less. However, electrolytic capacitors do have a much larger capacitance \[0.1F\] because of very minute separation between the electrolytic capacitors.)

Ans: Capacitance of a parallel capacitor, \[V=2F\]

Distance between the two plates, \[d=0.5cm=0.5\times {{10}^{-2}}m\]. Capacitance of a parallel plate capacitor is given by the relation,

\[{{\in }_{0}}=\]permittivity of free space\[=8.85\times {{10}^{-12}}{{C}^{2}}{{N}^{-1}}{{m}^{2}}\]

\[A=\frac{2\times 0.5\times {{10}^{-2}}}{8.85\times {{10}^{-12}}}\]

\[\Rightarrow A=1130k{{m}^{2}}\]

Hence, the area of the plates is too large. To avoid this situation, the capacitance is taken in the range of \[\mu F\].

21. A spherical capacitor consists of two concentric spherical conductors, held in position by suitable insulating supports (Fig. 2.36).

(Image will be uploaded soon)

Show that the capacitance of a spherical capacitor is given by\[C=\frac{4\pi {{\in }_{0}}{{r}_{1}}{{r}_{2}}}{{{r}_{1}}-{{r}_{2}}}\] where \[{{r}_{1}}\] and \[{{r}_{2}}\] re the radii of outer and inner spheres, respectively.

Radius of the outer shell is given in the question as\[={{r}_{1}}\]

Radius of the inner shell is given as\[={{r}_{2}}\]

The inner surface of the outer shell's charge can be given as \[+Q\].

\[V=\frac{Q}{4\pi {{\in }_{0}}{{r}_{2}}}-\frac{Q}{4\pi {{\in }_{0}}{{r}_{1}}}\]

The outer surface of the inner shell has induced charge \[-Q\]. Potential difference between the two shells is given by,

\[V=\frac{Q}{4\pi {{\in }_{0}}}\left[ \frac{1}{{{r}_{2}}}-\frac{1}{{{r}_{1}}} \right]\]

\[\Rightarrow V=\frac{Q\left( {{r}_{1}}-{{r}_{2}} \right)}{4\pi {{\in }_{0}}{{r}_{1}}{{r}_{2}}}\]

Capacitance of the given system is given by,

\[C=\frac{Charge\,\left( Q \right)}{Potential\,Difference\,\left( V \right)}\]

\[\Rightarrow C=\frac{4\pi {{\in }_{0}}{{r}_{1}}{{r}_{2}}}{{{r}_{1}}-{{r}_{2}}}\]

22. A cylindrical capacitor has two co-axial cylinders of length \[15cm\] and radii \[1.5cm\] and \[1.4cm\]. The outer cylinder is earthed and the inner cylinder is given a charge of \[3.5\mu C\]. Determine the capacitance of the system and the potential of the inner cylinder. Neglect end effects (i.e., bending of field lines at the ends).

Length of a co-axial cylinder is given as, \[l=15cm=0.15m\]

Radius of the outer cylinder is given as, \[{{r}_{1}}=1.5cm=0.015m\]

Radius of the inner cylinder is given as, \[{{r}_{2}}=1.4cm=0.014m\]

Charge on the inner cylinder, \[q=3.5\mu C=3.5\times {{10}^{-6}}C\]

Capacitance of a co-axial cylinder of radii \[{{r}_{1}}\] and \[{{r}_{2}}\] is given by the relation,

\[C=\frac{2\pi {{\in }_{0}}l}{{{\log }_{2}}\left( \frac{{{r}_{1}}}{{{r}_{2}}} \right)}\]

\[{{\in }_{0}}=\]permittivity of free space\[=8.85\times {{10}^{-12}}{{N}^{-1}}{{m}^{-2}}{{C}^{2}}\]

\[C=\frac{2\pi \times 8.85\times {{10}^{-12}}\times 0.15}{2.3026{{\log }_{10}}\left( \frac{0.15}{0.14} \right)}\]

\[\Rightarrow C=\frac{2\pi \times 8.85\times {{10}^{-12}}\times 0.15}{2.3026\times 0.0299}=1.2\times {{10}^{-10}}F\]

Potential difference of the inner cylinder is given by,

\[V=\frac{Q}{C}\]

\[\Rightarrow V=\frac{3.5\times {{10}^{-6}}}{1.2\times {{10}^{-10}}}=2.92\times {{10}^{4}}V\]

23. A parallel plate capacitor is to be designed with a voltage rating \[1kV\], using a material of dielectric constant \[3\]and dielectric strength about \[{{10}^{7}}V{{m}^{-1}}\]. (Dielectric strength is the maximum electric field a material can tolerate without breakdown, i.e., without starting to conduct electricity through partial ionisation). For safety, we should like the field never to exceed, say \[10%\] of the dielectric strength. What minimum area of the plates is required to have a capacitance of \[50pF\]?

Potential rating of a parallel plate capacitor, \[V=1kV=1000V\]

Dielectric constant of a material, \[{{\in }_{r}}=3\] Dielectric strength\[={{10}^{7}}V/m\]

For safety, the field intensity never exceeds 10%of the dielectric strength. Hence, electric field intensity, \[E=10%\] of \[{{10}^{7}}={{10}^{6}}V/m\]

Capacitance of the parallel plate capacitor, $C=50pF=50\times10^{-12}F$

Distance between the plates is given by,

\[d=\frac{V}{E}\]

\[\Rightarrow d=\frac{1000}{{{10}^{6}}}\]

\[\Rightarrow d={{10}^{-3}}m\]

Capacitance is given by the relation,

\[C=\frac{{{\in }_{0}}{{\in }_{1}}A}{d}\]

\[A=\]area of each plate

\[A=\frac{CD}{{{\in }_{0}}{{\in }_{1}}}\]

\[\Rightarrow A=\frac{50\times {{10}^{-12}}\times {{10}^{-3}}}{8.85\times {{10}^{-12}}\times 3}=19c{{m}^{2}}\]

Hence, the area of each plate is about \[19c{{m}^{2}}\].

Long Answer Questions (5 Mark Questions)

(a) Define dielectric constant in terms of the capacitance of a capacitor. On what factor does the capacitance of a parallel plate capacitor with dielectric depend?

Ans: Dielectric constant can be expressed as the ratio of capacitance of a capacitor when the dielectric is filled in between the plates to the capacitance of a capacitor when there is vacuum in between the plates.

In \[K=\frac{{{C}_{m}}}{{{C}_{o}}}=\frac{\text{Capacitance of a capacitor when dielectric is in between the plates}}{\text{Capacitance of a capacitor with vacuum in between the plates}}\]

Capacitance of a parallel plate capacitor with dielectric depends on the following factors

\[{{C}_{m}}=\frac{KA{{\in }_{0}}}{d}\]

Area of the plates

Distance between the plates

Dielectric constant of the dielectric between the plates.

(b) Find the ratio of the potential differences that must be applied across the 

(i) Parallel

(ii) Series combination of two identical capacitors so that the energy stored in the two cases becomes the same.

Ans: Let the capacitance of each capacitor be \[C\]

\[{{C}_{p}}=\frac{C\times C}{C+C}=\frac{C}{2}\] …… (in series)

Let \[{{V}_{p}}\] and \[{{V}_{s}}\] be the values of potential difference

Thus \[{{U}_{p}}=\frac{1}{2}\times {{C}_{p}}\times {{V}_{p}}^{2}=\frac{1}{2}\times 2C\times {{V}_{p}}^{2}=C{{V}_{p}}^{2}\]

\[{{U}_{s}}=\frac{1}{2}\times {{C}_{s}}\times {{V}_{s}}^{2}=\frac{1}{2}\times \frac{C}{2}\times {{V}_{s}}^{2}=\frac{C{{V}_{s}}^{2}}{4}\]

But \[{{U}_{p}}={{U}_{s}}\] …… (given)

\[C{{V}_{p}}^{2}=\frac{C{{V}_{s}}^{2}}{4}\]

\[\Rightarrow \frac{{{V}_{p}}^{2}}{{{V}_{s}}^{2}}=\frac{1}{4}\]

\[\Rightarrow {{V}_{p}}:{{V}_{s}}=1:2\]

(a) An air-filled capacitor is given a charge of \[2\mu C\] raising its potential to \[200V\]. If on inserting a dielectric medium, its potential falls to \[50V\], what is the dielectric constant of the medium?

Ans: \[K=\frac{V}{{{V}'}}=\frac{200}{50}=4\] (where \[V=200V\] for air filled capacitor \[{V}'=50\] after insertion of a dielectric)

(b) A conducting slab of thickness \[t\] is introduced without touching between plates of a parallel plate capacitor separated by a distance d , \[t<d\] . Derive an expression for the capacitance of a capacitor?

Ans: For a parallel plate capacitor when air/vacuum is in between the plates \[{{C}_{0}}=\frac{A{{\in }_{0}}}{d}\]

Since the electric field inside a conducting stab is zero, hence the electric field exists only between the space \[\left( d-t \right)\Rightarrow V={{E}_{0}}\left( d-t \right)\]

Where \[{{E}_{0}}\] is the electric field between the plates.

And \[{{E}_{0}}=\frac{\sigma }{{{\in }_{0}}}=\frac{q}{A{{\in }_{0}}}\]

Where \[A\] is the area of each plate

\[\Rightarrow V=\frac{q}{A{{\in }_{0}}}\left( d-t \right)\]

Hence capacitor of a parallel plate capacitor

\[C=\frac{q}{V}=\frac{q}{q\left( d-t \right)}A{{\in }_{0}}\]

\[\Rightarrow C=\frac{A{{\in }_{0}}}{d-t}\]

\[\Rightarrow C=\frac{A{{\in }_{0}}}{d\left( 1-\frac{t}{d} \right)}\]

\[\Rightarrow C=\frac{{{C}_{0}}}{\left( 1-\frac{t}{d} \right)}\]

3. Figure (a) and (b) shows the field lines of a single positive and negative changes respectively

(a) Give the signs of the potential: \[{{V}_{P}}-{{V}_{Q}}\] and \[{{V}_{B}}-{{V}_{A}}\]

Ans: We know the relation as \[V\propto \frac{1}{r}\]

\[{{V}_{P}}>{{V}_{Q}}\Rightarrow {{V}_{P}}-{{V}_{Q}}\,=\,Positive\]

\[{{V}_{A}}<{{V}_{B}}\Rightarrow {{V}_{B}}-{{V}_{A}}\,=\,Positive\]

Because \[{{V}_{B}}\] is less negative than \[{{V}_{A}}\].

(b) Give the sign of the work done by the field in moving a small positive change from \[Q\] to \[P\].

Ans: In moving a positive change form \[Q\] to \[P\] work has to be done against the electric field so it is negative, and the sign is, hence, negative.

(c) Give the sign of the work done by the field in moving a small negative change from \[B\] to \[A\].

Ans:  In moving a negative change form \[B\] to \[A\] work is done along the same direction of the field so it is positive, and the sign is, hence, positive.

4. With the help of a labelled diagram, explain the principle, construction and working of a Van de Graff generator. Mention its applications.

Ans: Van de Graff generator can be expressed as a device which is capable of producing a high potential of the order of million volts.

Principle: 

The charge always resides on the outer surface of the hollow conductor.

The electric discharge in air or gas takes place readily at the pointed ends of the conductors.

Construction of Van de Graff Generator: 

It consists of a large hollow metallic sphere \[S\] mounted on two insulating columns A and B and an endless belt made up of rubber which is running over two pulleys \[{{P}_{1}}\] and \[{{P}_{2}}\] with the help of an electric motor. \[{{B}_{1}}\] and \[{{B}_{2}}\] are two sharp metallic brushes. The lower brush \[{{B}_{1}}\] is given a positive potential by high tension battery and is called a spray brush while the upper brush \[{{B}_{2}}\] connected to the inner part of the sphere \[S\].

Working: 

When brush \[{{B}_{1}}\] is given a higher positive potential then it produces ions, due to the action of sharp points. Thus, the positive ions so produced get sprayed on the belt due to repulsion between positive ions and the positive charge on brush \[{{B}_{1}}\]. Then it is carried upward by the moving belt. The pointed end of \[{{B}_{2}}\] just touches the belt collects the positive change and make it move to the outer surface of the sphere S. This process continues and the potential of the shell rises to several million volts.

Applications: 

Particles like protons, deuterons, \[\alpha \]-particles etc are accelerated to high speeds and energies.

5. Two charges \[5\times {{10}^{-8}}C\] and \[-3\times {{10}^{-8}}C\] are located \[16cm\] apart. At what point(s) on the line joining the two charges is the electric potential zero? Take the potential at infinity to be zero.

Ans: There are two charges which are given as,

\[{{q}_{1}}=5\times {{10}^{-8}}C\]

\[{{q}_{2}}=-3\times {{10}^{-8}}C\]

Distance between the two charges, \[d=16cm=0.16m\]

Consider a point \[P\] on the line joining the two charges, as shown in the given figure.

\[r=\] distance of point \[P\] from charge \[{{q}_{1}}\]

Let the electric potential \[V\] at point \[P\] be zero.

Potential at point \[P\] is the sum of potentials caused by charges \[{{q}_{1}}\] and \[{{q}_{2}}\] respectively.

\[V=\frac{{{q}_{1}}}{4\pi {{\in }_{0}}r}+\frac{{{q}_{2}}}{4\pi {{\in }_{0}}\left( d-r \right)}\] …… (1)

For \[V=0\], equation (1) reduces to

\[\frac{{{q}_{1}}}{4\pi {{\in }_{0}}r}=-\frac{{{q}_{2}}}{4\pi {{\in }_{0}}\left( d-r \right)}\]

\[\Rightarrow \frac{{{q}_{1}}}{r}=-\frac{{{q}_{2}}}{\left( d-r \right)}\]

\[\Rightarrow \frac{5\times {{10}^{-8}}}{r}=\frac{\left( -3\times {{10}^{-8}} \right)}{\left( 0.16-r \right)}\]

\[\Rightarrow \frac{0.16}{r}=\frac{8}{5}\]

\[\Rightarrow r=0.1m=10cm\]

Therefore, the potential is zero at a distance of \[10cm\] from the positive charge between the charges.

Suppose point \[P\] is outside the system of two charges where potential is zero, as shown in the given figure.

For this arrangement, potential is given

 \[V=\frac{{{q}_{1}}}{4\pi {{\in }_{0}}s}+\frac{{{q}_{2}}}{4\pi {{\in }_{0}}\left( s-d \right)}\]…… (2)

For \[V=0\], equation (2) reduces to

\[\frac{{{q}_{1}}}{4\pi {{\in }_{0}}s}=-\frac{{{q}_{2}}}{4\pi {{\in }_{0}}\left( s-d \right)}\]

\[\Rightarrow \frac{{{q}_{1}}}{s}=-\frac{{{q}_{2}}}{\left( s-d \right)}\]

\[\Rightarrow \frac{5\times {{10}^{-8}}}{s}=\frac{\left( -3\times {{10}^{-8}} \right)}{\left( s-0.16 \right)}\]

\[\Rightarrow 1-\frac{0.16}{s}=\frac{3}{5}\]

\[\Rightarrow \frac{0.16}{s}=\frac{2}{5}\]

\[\Rightarrow s=0.4m=40cm\]

Therefore, the potential is zero at a distance of \[40cm\] from the positive charge outside the system of charges.

6. A parallel plate capacitor with air between the plates has a capacitance of \[8pF\], \[\left( 1pF={{10}^{-12}}F \right)\]. What will be the capacitance if the distance between the plates is reduced by half, and the space between them is filled with a substance of dielectric constant \[6\]?

Ans: Capacitance between the parallel plates of the capacitor, \[C=8pF\]

Initially, the distance between the parallel plates was \[d\] and it was filled with air. Dielectric constant of air, \[k=1\]

Capacitance, \[C\] is given by the formula,

\[C=\frac{kA{{\in }_{0}}}{d}\]

\[\Rightarrow C=\frac{A{{\in }_{0}}}{d}\] ……(1)

\[A=\] Area of each plate

\[{{\in }_{0}}=\] Permittivity of free space

If distance between the plates is reduced to half, then the new distance can be given as, \[{{d}^{-TM}}=\frac{d}{2}\]. Dielectric constant of the substance filled in between the plates, \[{k}'=6\]. 

Therefore, capacitance of the capacitor becomes

\[{C}'=\frac{{k}'A{{\in }_{0}}}{d}=\frac{6A{{\in }_{0}}}{\frac{d}{2}}\] …… (2)

Taking the ratios of equation \[1\] and \[2\], we obtain

\[{C}'=2\times 6C\]

\[\Rightarrow {C}'=12C\]

\[\Rightarrow {C}'=12\times 8=96pF\]

Therefore, the capacitance between the plates is \[96pF\].

7. A charge of \[8mC\] is located at the origin. Calculate the work done in taking a small charge of \[-2\times {{10}^{-9}}C\] from a point \[P\left( 0,0,3 \right)cm\] to a point \[Q\left( 0,4,0 \right)cm\], via a point \[R\left( 0,6,9 \right)cm\] .

Ans: Charge located at the origin is given as, \[q=8mC=8\times {{10}^{3}}C\].

Magnitude of a small charge, which is taken from a point \[P\] to point \[R\] to point \[Q\].

\[{{q}_{1}}=-2\times {{10}^{-9}}C\]

All the points are represented in the figure below.

Point \[P\] is at a distance, \[{{d}_{1}}=3cm\] from the origin along z-axis. Point \[Q\] is at a distance, \[{{d}_{2}}=4cm\], from the origin along y-axis.

Potential at point \[P\]can be given as, \[{{V}_{1}}=\frac{q}{4\pi {{\in }_{0}}{{d}_{1}}}\]

Potential at point \[Q\], \[{{V}_{2}}=\frac{q}{4\pi {{\in }_{0}}{{d}_{2}}}\]

Work done \[W\] by the electrostatic force is independent of the path.

\[W={{q}_{1}}\left( {{V}_{2}}-{{V}_{1}} \right)\]

\[\Rightarrow W={{q}_{1}}\left[ \frac{q}{4\pi {{\in }_{0}}{{d}_{2}}}-\frac{q}{4\pi {{\in }_{0}}{{d}_{1}}} \right]\]

\[\Rightarrow W=\frac{{{q}_{1}}{{q}_{2}}}{4\pi {{\in }_{0}}}\left[ \frac{1}{{{d}_{2}}}-\frac{1}{{{d}_{1}}} \right]\] …… (1)

Where, 

\[\frac{1}{4\pi {{\in }_{0}}}=9\times {{10}^{9}}N{{m}^{2}}{{C}^{-2}}\]

\[\Rightarrow W=9\times {{10}^{9}}\times 8\times {{10}^{-3}}\times \left( -2\times {{10}^{-9}} \right)\left[ \frac{1}{0.04}-\frac{1}{0.03} \right]\]

\[\Rightarrow W=-144\times {{10}^{-3}}\times \left( \frac{-25}{3} \right)\]

\[\Rightarrow W=1.27J\]

Therefore, work done during the process is \[1.27J\].

8. A cube of side \[b\]has a charge \[q\] at each of its vertices. Determine the potential and electric field due to this charge array at the centre of the cube.

Ans: Length of the side of a cube is given as \[=b\]

Charge at each of its vertices\[=q\]

A cube of side \[b\] is shown in the following figure.

\[d=\]Diagonal of one of the six faces of the cube

\[\Rightarrow {{d}^{2}}=\sqrt{{{b}^{2}}+{{b}^{2}}}=\sqrt{2{{b}^{2}}}\]

\[\Rightarrow d=b\sqrt{2}\]

\[l=\] length of the diagonal of the cube

\[{{l}^{2}}=\sqrt{{{d}^{2}}+{{b}^{2}}}\]

\[\Rightarrow {{l}^{2}}=\sqrt{{{\left( \sqrt{2}b \right)}^{2}}+{{b}^{2}}}\]

\[\Rightarrow {{l}^{2}}=\sqrt{2{{b}^{2}}+{{b}^{2}}}\]

\[\Rightarrow {{l}^{2}}=\sqrt{3{{b}^{2}}}\]

\[\Rightarrow l=b\sqrt{3}\]

\[\Rightarrow r=\frac{l}{2}=\frac{b\sqrt{2}}{2}\] is the difference between the centre of the cube and one of the eight vertices.

The electric potential \[V\] at the centre of the cube is due to the presence of eight charges at the vertices.

\[V=\frac{8q}{4\pi {{\in }_{0}}}\]

\[\Rightarrow V=\frac{8q}{4\pi {{\in }_{0}}\left( \frac{b\sqrt{3}}{2} \right)}\]

\[\Rightarrow V=\frac{4q}{\sqrt{3}\pi {{\in }_{0}}b}\]

Therefore, the potential at the centre of the cube is \[\frac{4q}{\sqrt{3}\pi {{\in }_{0}}b}\].

The electric field at the centre of the cube, due to the eight charges, gets cancelled. This is because the charges are distributed symmetrically with respect to the cube. 

Therefore, the electric field is zero at the centre.

9. Two tiny spheres carrying charges \[1.5\mu C\] and \[2.5\mu C\] are located \[30cm\] apart. Find the potential and electric field:

(a) At the mid-point of the line joining the two charges, and

Ans: The charges are depicted as:

Two charges placed at points \[A\] and \[B\] are depicted in the provided figure. O is the mid-point of the line joining the two charges.

Magnitude of charge located at \[A\], \[{{q}_{1}}=1.5\mu C\]

Magnitude of charge located at \[B\], \[{{q}_{2}}=2.5\mu C\]

Distance between the two charges, \[d=30cm=0.3m\]

Let \[{{V}_{1}}\]and  \[{{E}_{1}}\] are electric potential and electric field respectively at O.

\[{{V}_{1}}=\] Potential due to charge at \[A+\]Potential due to charge at \[B\]

\[{{V}_{1}}=\frac{{{q}_{1}}}{4\pi {{\in }_{0}}\left( \frac{d}{2} \right)}+\frac{{{q}_{2}}}{4\pi {{\in }_{0}}\left( \frac{d}{2} \right)}=\frac{1}{4\pi {{\in }_{0}}\left( \frac{d}{2} \right)}\left( {{q}_{1}}+{{q}_{2}} \right)\]

Where, \[\frac{1}{4\pi {{\in }_{0}}}=9\times {{10}^{9}}N{{C}^{2}}{{m}^{-2}}\]

\[{{V}_{1}}=\frac{9\times {{10}^{9}}\times {{10}^{-5}}}{\left( \frac{0.30}{2} \right)}\left( 2.5+1.5 \right)\]

\[{{V}_{1}}=2.4\times {{10}^{5}}V\]

\[{{E}_{1}}=\] Electric field due to \[{{q}_{2}}\] \[-\]Electric field due to \[{{q}_{1}}\]

\[\Rightarrow {{E}_{1}}=\frac{{{q}_{2}}}{4\pi {{\in }_{0}}{{\left( \frac{d}{2} \right)}^{2}}}-\frac{{{q}_{1}}}{4\pi {{\in }_{0}}{{\left( \frac{d}{2} \right)}^{2}}}\]

\[\Rightarrow {{E}_{1}}=\frac{9\times {{10}^{9}}}{{{\left( \frac{0.30}{2} \right)}^{2}}}\times {{10}^{6}}\times \left( 2.5-1.5 \right)\]

\[\Rightarrow {{E}_{1}}=4\times {{10}^{5}}V{{m}^{-1}}\]

Therefore, the potential at mid-point is \[2.4\times {{10}^{5}}V\] and the electric field at mid-point is \[4\times {{10}^{5}}V{{m}^{-1}}\]. The field is directed from the larger charge to the smaller charge

(b) At a point \[10cm\] from this midpoint in a plane normal to the line passing through the mid-point.

Ans: Consider a point \[Z\]such that the normal distance \[OZ=10cm=0.1m\] as shown in the following figure.

\[{{V}_{2}}\] and \[{{E}_{2}}\] are the electric potential and electric field respectively at \[Z\]. It can be observed from the figure that distance,

\[BZ=AZ=\sqrt{{{\left( 0.1 \right)}^{2}}+{{\left( 0.15 \right)}^{2}}}=0.18m\]

\[{{V}_{2}}=\] Electric potential due to \[A+\]Electric potential due to point \[B\]

\[\Rightarrow {{V}_{2}}=\frac{{{q}_{2}}}{4\pi {{\in }_{0}}\left( AZ \right)}+\frac{{{q}_{1}}}{4\pi {{\in }_{0}}\left( BZ \right)}\]

\[\Rightarrow {{V}_{2}}=\frac{9\times {{10}^{9}}\times {{10}^{-5}}}{0.18}\left( 1.5+2.5 \right)\]

\[\Rightarrow {{V}_{2}}=2\times {{10}^{5}}V\]

Electric field due to \[{{q}_{1}}\] at \[Z\],

\[{{E}_{A}}=\frac{{{q}_{1}}}{4\pi {{\in }_{0}}\left( AZ \right)}\]

\[\Rightarrow {{E}_{A}}=\frac{9\times {{10}^{9}}\times 1.5\times {{10}^{-5}}}{0.18}\]

\[\Rightarrow {{E}_{A}}=0.416\times {{10}^{6}}V/m\]

Electric field due to \[{{q}_{2}}\] at \[Z\],

\[\Rightarrow {{E}_{B}}=\frac{9\times {{10}^{9}}\times 2.5\times {{10}^{-5}}}{{{\left( 0.18 \right)}^{2}}}\]

\[\Rightarrow {{E}_{B}}=0.69\times {{10}^{6}}V/m\]

The resultant field intensity at \[Z\]

\[E=\sqrt{{{E}^{2}}_{A}+{{E}^{2}}_{B}+2{{E}_{A}}{{E}_{B}}\cos 2\theta }\]

Where, \[2\theta \] is the angle, \[\angle AZB\]

From the given figure, we obtain

\[\Rightarrow \theta ={{\cos }^{-1}}0.5556={{56.25}^{\circ }}\]

\[E=\sqrt{{{\left( 0.416\times {{10}^{6}} \right)}^{2}}\times 2\times 0.416\times 0.69\times {{10}^{12}}\times \left( -0.38 \right)}\]

\[\Rightarrow E=6.6\times {{10}^{5}}V{{m}^{-1}}\]

Therefore, the potential at a point \[10cm\] (perpendicular to the mid-point) is $ 2.0 \times {{10}^{5}} V$ and the electric field is \[6.6\times {{10}^{5}}V{{m}^{-1}}\].

(a) Show that the normal component of electrostatic field has a discontinuity from one side of a charged surface to another given by \[\left( \overline{{{E}_{2}}}-\overline{{{E}_{1}}} \right)\cdot \hat{n}=\frac{\sigma }{{{\in }_{0}}}\]. Where \[\hat{n}\] is a unit vector normal to the surface at a point and \[\sigma \] is the surface charge density at that point. (The direction of \[\hat{n}\] is from side \[1\]to side \[2\]). Hence show that just outside a conductor, the electric field is \[\frac{{\hat{n}}}{{{\in }_{0}}}\].

Ans: Electric field on one side of a charged body is \[{{E}_{1}}\] and electric field on the other side of the same body is \[{{E}_{2}}\]. If an infinite plane charged body has a uniform thickness, then electric field due to one surface of the charged body is given by,

\[\overline{{{E}_{1}}}=-\frac{\sigma }{2{{\in }_{0}}}\hat{n}\]

\[\hat{n}=\] Unit vector normal to the surface at a point 

\[\sigma =\]Surface charge density at that point 

Electric field due to the other surface of the charge body,

\[\overline{{{E}_{2}}}=-\frac{\sigma }{2{{\in }_{0}}}\hat{n}\]

Electric field at any point due to the two surfaces,

$\left( \overline{{{E}_{2}}}-\overline{{{E}_{1}}} \right)=\frac{\sigma }{2{{\in }_{0}}}\hat{n}+\frac{\sigma }{2{{\in }_{0}}}\hat{n}+\frac{\sigma }{{{\in }_{0}}}\hat{n} $

$\Rightarrow \left( \overline{{{E}_{2}}}-\overline{{{E}_{1}}} \right)\cdot \hat{n}=\frac{\sigma }{{{\in }_{0}}} $

Since inside a closed conductor, \[\overline{{{E}_{1}}}=0\],

\[\overline{E}=\overline{{{E}_{2}}}=-\frac{\sigma }{2{{\in }_{0}}}\hat{n}\]

Therefore, the electric field just outside the conductor is \[\frac{\sigma }{{{\in }_{0}}}\hat{n}\].

(b) Show that the tangential component of an electrostatic field is continuous from one side of a charged surface to another. 

Ans: When a charge particle is moved from one point to the other on a closed loop, the work done by the electrostatic field is zero. Hence, the tangential component of an electrostatic field is continuous from one side of a charged surface to the other.

11. A long charged cylinder of linear charge density \[\lambda \] is surrounded by a hollow co-axial conducting cylinder. What is the electric field in the space between the two cylinders?

Ans: Charge density of the long, charged cylinder of length \[L\] and radius \[r\] is \[\lambda \].

Another cylinder of the same length surrounds the previous cylinder. The radius of this cylinder is \[R\].

Let \[E\] be the electric field produced in the space between the two cylinders.

Electric flux through the Gaussian surface is given by Gauss's theorem as, \[\phi =E\left( 2\pi d \right)L\]

\[d=\]Distance of a point from the common axis of the cylinders. Let \[q\]be the total charge on the cylinder.

It can be written as

\[\Rightarrow \phi =E\left( 2\pi d \right)L=\frac{q}{{{\in }_{0}}}\]

\[q=\] Charge on the inner sphere of the outer cylinder

\[\Rightarrow E\left( 2\pi d \right)L=\frac{\lambda L}{{{\in }_{0}}}\]

\[\Rightarrow E=\frac{\lambda }{2\pi {{\in }_{0}}d}\]

Therefore, the electric field in the space between the two cylinders is \[\frac{\lambda }{2\pi {{\in }_{0}}d}\].

12. In a hydrogen atom, the electron and proton are bound at a distance of about \[0.53{{A}^{\circ }}\].

(a) Estimate the potential energy of the system in \[eV\], taking the zero of the potential energy at infinite separation of the electron from the proton.

Ans: The distance between the electron-proton of a hydrogen atom is given as, \[d=0.53\overset{{}^\circ }{\mathop{A}}\,\].

Charge on an electron, \[{{q}_{1}}=1.6\times {{10}^{-19}}C\]

Charge on a proton, \[{{q}_{2}}=+1.6\times {{10}^{-19}}C\]

Potential at infinity is zero.

\[=0-\frac{{{q}_{1}}{{q}_{2}}}{4\pi {{\in }_{0}}d}\]

Potential energy of the system, \[p-e=\] Potential energy at infinity – Potential energy t distance, \[d\]

\[{{\in }_{0}}\] is the permittivity of free space

We can further express,

\[\Rightarrow \frac{1}{4\pi {{\in }_{0}}}=0-\frac{9\times {{10}^{9}}\times {{\left( 1.6\times {{10}^{-19}} \right)}^{2}}}{0.53\times {{10}^{10}}}\]

\[Potential\,Energy=\frac{-43.7\times {{10}^{-19}}}{1.6\times {{10}^{-19}}}\]

\[\Rightarrow Potential\,Energy=-27.2eV\]

Therefore, the potential energy of the system is \[-27.2eV\].

(b) What is the minimum work required to free the electron, given that its kinetic energy in the orbit is half the magnitude of potential energy obtained in (a)

Ans: Kinetic energy can be given as half times the magnitude of potential energy.

Kinetic Energy\[=\frac{1}{2}\times \left( -27.2eV \right)=13.6eV\]

Total energy \[=13.6eV\]

Therefore, the minimum work required to free the electron is \[13.6eV\].

(c) What are the answers to (a) and (b) above if the zero of potential energy is taken at \[1.06\overset{{}^\circ }{\mathop{A}}\,\] separation?

Ans: When zero of potential energy is taken, \[{{d}_{1}}=1.06\overset{{}^\circ }{\mathop{A}}\,\]

$\therefore$ Potential energy of the system = Potential energy at \[{{d}_{1}}\]\[-\]Potential energy at \[d\].

\[Potential\,Energy\,of\,the\,\text{ }system=\frac{{{q}_{1}}{{q}_{2}}}{4\pi {{\in }_{0}}{{d}_{1}}}-27.2eV\]

\[\Rightarrow Potential\,Energy\,of\,the\,\text{ }system=\frac{9\times {{10}^{9}}\times {{\left( 1.6\times {{10}^{-19}} \right)}^{2}}}{1.06\times {{10}^{-10}}}-27.2eV\]

\[\Rightarrow 21.73\times {{10}^{-10}}J-27.2eV\]

\[\Rightarrow 13.58eV-27.2eV\]

\[\Rightarrow 13.6eV\]

13. Two charges \[-q\] and \[+q\] are located at points \[\left( 0,0,-a \right)\] and \[\left( 0,0,a \right)\] , respectively.

(a) What is the electrostatic potential at these points?

Ans:  In this situation, charge \[-q\] is located at \[\left( 0,0,-a \right)\] and charge \[+q\] is located at \[\left( 0,0,a \right)\]. Therefore, they form a dipole. Point \[\left( 0,0,z \right)\] is on the axis of this dipole and point \[\left( x,y,0 \right)\] is normal to the axis of the dipole. Hence, the electrostatic potential at point \[\left( x,y,0 \right)\] is zero. Electrostatic potential at point \[\left( 0,0,z \right)\] is given by,

\[V=\frac{1}{4\pi {{\in }_{0}}}\left( \frac{q}{z-a} \right)+\frac{1}{4\pi {{\in }_{0}}}\]

\[\Rightarrow V=\frac{q\left( z+a-z+a \right)}{4\pi {{\in }_{0}}\left( {{z}^{2}}-{{a}^{2}} \right)}\]

\[\Rightarrow V=\frac{2qa}{4\pi {{\in }_{0}}\left( {{z}^{2}}-{{a}^{2}} \right)}\]

\[\Rightarrow V=\frac{p}{4\pi {{\in }_{0}}\left( {{z}^{2}}-{{a}^{2}} \right)}\]

\[p=\]Dipole moment of the system of two charges \[=2qa\]

(b) Obtain the dependence of potential on the distance \[r\] of a point from the origin when \[\frac{r}{a}>>1\].

Ans: Distance \[r\] is much greater than half of the distance between the two charges. Therefore, the potential \[\left( V \right)\] at a distance \[r\] is inversely proportional to the square of the distance.

i.e., \[4\propto \frac{1}{{{r}^{2}}}\]

(c) How much work is done in moving a small test charge from the point \[1.06{{A}^{\circ }}\] to \[\left( -7,0,0 \right)\] along the x-axis? Does the answer change if the path of the test charge between the same points is not along the x-axis?

Ans:   Zero

The answer does not change if the path of the test charge is not along the x-axis.

A test charge is moved from point \[\left( 5,0,0 \right)\] to point \[\left( -7,0,0 \right)\] along the x-axis. Electrostatic potential \[{{V}_{1}}\] at point \[\left( 5,0,0 \right)\] is given by,

\[{{V}_{1}}=\frac{-q}{4\pi {{\in }_{0}}}\frac{1}{\sqrt{{{\left( 5-0 \right)}^{2}}-{{\left( -a \right)}^{2}}}}+\frac{q}{4\pi {{\in }_{0}}}\frac{1}{\sqrt{{{\left( 5-0 \right)}^{2}}-{{\left( -a \right)}^{2}}}}\]

\[\Rightarrow {{V}_{1}}=\frac{-q}{4\pi {{\in }_{0}}\sqrt{25+{{a}^{2}}}}+\frac{q}{4\pi {{\in }_{0}}\sqrt{25+{{a}^{2}}}}\]

\[\Rightarrow {{V}_{1}}=0\]

Electrostatic potential, \[{{V}_{2}}\] at point \[\left( -7,0,0 \right)\] is given by,

\[{{V}_{2}}=\frac{-q}{4\pi {{\in }_{0}}}\frac{1}{\sqrt{{{\left( 7 \right)}^{2}}-{{\left( -a \right)}^{2}}}}+\frac{q}{4\pi {{\in }_{0}}}\frac{1}{\sqrt{{{\left( -7 \right)}^{2}}-{{\left( -a \right)}^{2}}}}\]

\[\Rightarrow {{V}_{2}}=\frac{-q}{4\pi {{\in }_{0}}\sqrt{49+{{a}^{2}}}}+\frac{q}{4\pi {{\in }_{0}}\sqrt{49+{{a}^{2}}}}\]

\[\Rightarrow {{V}_{2}}=0\]

Therefore, no work is done in moving a small test charge from point \[\left( 5,0,0 \right)\] to point \[\left( -7,0,0 \right)\] along the x-axis.

The answer will not change because work done by the electrostatic field in moving a test charge between the two points is not dependent of the path connecting the two points.

14. Figure 2.34 shows a charge array known as an electric quadrupole. For a point on the axis of the quadrupole, obtain the dependence of potential on \[r\] for \[\frac{r}{a}>>1\] and contrast your results with that due to an electric dipole, and an electric monopole (i.e., a single charge)

Ans: Four charges of the same magnitude are placed at points \[X\], \[Y\] and \[Z\] respectively, as depicted in the following figure.

A point is located at \[P\], which is \[r\] distance away from point \[Y\]. The system of charges forms an electric quadrupole.

It can be considered that the system of the electric quadrupole has three charges

Charge \[+q\] placed at point \[X\]

Charge \[-2q\] placed at point \[Y\]

Charge \[+q\] placed at point \[Z\]

\[XY=YZ=a\]

Electrostatic potential that is caused by the system of three charges at point \[P\] can be given by,

\[V=\frac{1}{4\pi {{\in }_{0}}}\left[ \frac{q}{XP}-\frac{2q}{YP}+\frac{q}{ZP} \right]\]

\[\Rightarrow V=\frac{1}{4\pi {{\in }_{0}}}\left[ \frac{q}{r+a}-\frac{2q}{r}+\frac{q}{r-a} \right]\]

\[\Rightarrow V=\frac{q}{4\pi {{\in }_{0}}}\left[ \frac{r\left( r-a \right)-2\left( r+a \right)\left( r-a \right)+r\left( r+a \right)}{r\left( r+a \right)\left( r-a \right)} \right]\]

\[\Rightarrow V=\frac{q}{4\pi {{\in }_{0}}}\left[ \frac{{{r}^{2}}-ra-2{{r}^{2}}+2{{a}^{2}}+{{r}^{2}}+ra}{r\left( {{r}^{2}}-{{a}^{2}} \right)} \right]\]

\[\Rightarrow V=\frac{q}{4\pi {{\in }_{0}}}\left[ \frac{2{{a}^{2}}}{r\left( {{r}^{2}}-{{a}^{2}} \right)} \right]\]

\[\Rightarrow V=\frac{2q{{a}^{2}}}{4\pi {{\in }_{0}}{{r}^{3}}\left( 1-\frac{{{a}^{2}}}{{{r}^{2}}} \right)}\]

Since \[\frac{r}{a}>>1\]

\[\frac{{{r}^{2}}}{{{a}^{2}}}\] is taken as negligible.

\[V=\frac{2q{{a}^{2}}}{4\pi {{\in }_{0}}{{r}^{3}}}\]

It can be inferred that potential, \[V\propto \frac{1}{{{r}^{3}}}\]

But it is known that for a dipole, \[V\propto \frac{1}{{{r}^{2}}}\]

And, for a monopole, \[V\propto \frac{1}{r}\]

15. An electrical technician requires a potential difference of $2\mu F$ in a circuit across a potential difference of  \[1kV\]. A large number of  $2\mu F$ capacitors are available to him, each of which can withstand a potential difference of not more than $400V$. Suggest a possible arrangement that requires the minimum number of capacitors.

Ans: Total required capacitance is given as, $C=2\mu F$ 

Potential difference is given as, $V=1kV=1000V$

Capacitance of each capacitor, ${{C}_{1}}=1\mu F$

Each capacitor can hold a potential difference,

 ${{V}_{1}}=400V$

$\frac{1000}{400}=2.5$

Let us suppose a number of capacitors are connected in series and these series and these series circuits are connected in parallel (row) to each other. The potential difference across each row must be $1000V$ and the potential difference across each capacitor must be $400V$. Hence, the number of capacitors in each row is given as

Therefore, there are three capacitors in each row. 

Capacitance of each row$=\frac{1}{1+1+1}=\frac{1}{3}\mu F$

$\frac{1}{3}+\frac{1}{3}+\frac{1}{3}+.....\text{n terms = }\frac{n}{3}$

However, capacitance of the circuit is given as $2\mu F$.

$\therefore \frac{n}{3}=2$

Let there be $n$ rows, each having three capacitors, which are connected in parallel. Hence, equivalent capacitance of the circuit is given as

Hence, $6$ rows of three capacitors are present in the circuit. A minimum of $6\times 3$ i.e.,$18$ capacitors are required for the given arrangement.

16. Obtain the equivalent capacitance of the network in Fig. 2.35. For a $300V$ supply, determine the charge and voltage across each capacitor.

Ans: Capacitance of capacitor ${{C}_{1}}$ is given as $100pF$.

Capacitance of capacitor ${{C}_{2}}$ is given as $200pF$.

Capacitance of capacitor ${{C}_{3}}$ is given as $200pF$.

Capacitance of capacitor ${{C}_{4}}$ is given as $200pF$.

Supply potential, $V=300V$

Capacitors ${{C}_{2}}$ and ${{C}_{3}}$ are connected in series. Let their equivalent capacitance be C 1

$\therefore \frac{1}{{{C}^{1}}}=\frac{1}{200}+\frac{1}{200}=\frac{2}{200}\Rightarrow {{C}^{1}}=100pF$

Capacitors ${{C}^{1}}$ and $C'$ are parallel. Let their equivalent capacitance be ${{C}^{n}}$

$\therefore {{C}^{n}}={{C}^{1}}+{{C}_{1}}=100+100=200pF$

${{C}^{n}}$ and ${{C}_{4}}$ are connected in series. Let their equivalent capacitance be $C$.

$\therefore \frac{1}{C}=\frac{1}{{{C}^{n}}}+\frac{1}{{{C}_{4}}}=\frac{1}{200}+\frac{1}{100}=\frac{2+1}{200}\Rightarrow C=\frac{200}{3}pF$

Hence, the equivalent capacitance of the circuit is $\frac{200}{3}pF$.

 Potential difference across

${{C}^{n}}={V}''$

Potential difference across ${{C}_{4}}={{V}_{4}}$

$\therefore {{V}^{5}}+{{V}_{4}}=V=300V$

Charge on ${{C}_{1}}$ is given by,

${{Q}_{4}}=CV=\frac{200}{3}\times {{10}^{-12}}\times 300=2\times {{10}^{-8}}C$

$\therefore {{V}_{4}}=\frac{{{Q}_{4}}}{{{C}_{4}}}=\frac{2\times {{10}^{-8}}}{100\times {{10}^{-12}}}=200V$

$\therefore$ Voltage across ${{C}_{1}}$ is given by,

${{V}_{1}}=V-{{V}_{4}}=300-200=100V$

Hence, the potential difference, ${{V}_{1}}$ across ${{C}_{1}}$ is $100V$. Charge on ${{C}_{1}}$ is given by, 

${{Q}_{1}}={{C}_{1}}{{V}_{1}}=100\times {{10}^{-12}}\times 100={{10}^{-8}}$

Now,  ${{C}_{2}}$ and ${{C}_{3}}$ having the same capacitance have a potential difference of $100V$together. Since ${{C}_{2}}$ and ${{C}_{3}}$ are in series, the potential difference across ${{C}_{2}}$ and ${{C}_{3}}$ is given by,

${{V}_{2}}={{V}_{3}}=50V$

Therefore, charge on ${{C}_{2}}$ is given by,

${{Q}_{2}}={{C}_{2}}{{V}_{2}}=200\times {{10}^{-12}}\times 50={{10}^{-8}}C$

And charge on ${{C}_{3}}$ is given by,

${{Q}_{3}}={{C}_{3}}{{V}_{3}}=200\times {{10}^{-12}}\times 50={{10}^{-8}}C$

Therefore, the equivalent capacitance of the given circuit is  $\frac{200}{3}\text{pF with}$

${{Q}_{1}}={{10}^{-8}}C{{V}_{1}}=100V$

${{Q}_{2}}={{10}^{-8}}C{{V}_{2}}=50V$

${{Q}_{3}}={{10}^{-8}}C{{V}_{3}}=50V$

${{Q}_{4}}=2\times {{10}^{-8}}C{{V}_{4}}=200V$

17. The plates of a parallel plate capacitor have an area of $90c{{m}^{2}}$ each and are separated by $2.5mm$. The capacitor is charged by connecting it to a $400V$ supply.

(a) How much electrostatic energy is stored by the capacitor?

Ans: Area of the plates of a parallel plate capacitor is given as, $A=90c{{m}^{2}}=90\times {{10}^{-4}}{{m}^{2}}$

Distance between the plates, $d=2.5mm=2.5\times {{10}^{-3}}m$

Potential difference across the plates, $V=400V$

Capacitance of the capacitor is given by the relation,

$C=\frac{A{{\varepsilon }_{0}}}{d}$

Electrostatic energy stored in the capacitor is given by the relation, ${{E}_{1}}=\frac{1}{2}C{{V}^{2}}$

$=\frac{1}{2}\frac{{{\varepsilon }_{0}}A}{d}{{V}^{2}}$

${{\varepsilon }_{0}}=$ Permittivity of free space$=8.85\times {{10}^{-12}}{{C}^{2}}{{N}^{-1}}{{m}^{-2}}$

$\therefore {{E}_{1}}=\frac{1\times 8.85\times {{10}^{-12}}\times 90\times {{10}^{-4}}\times {{\left( 400 \right)}^{2}}}{2\times 2.5\times {{10}^{-3}}}=2.55\times {{10}^{-6}}J$

Hence, the electrostatic energy stored by the capacitor is $2.55\times {{10}^{-6}}J$.

(b) View this energy as stored in the electrostatic field between the plates, and obtain the energy per unit volume $u$. Hence arrive at a relation between $u$ and the magnitude of electric field $E$ between the plates.

Ans: Volume of the given capacitor,

${{V}^{1}}=A\times d=90\times {{10}^{-4}}\times 25\times {{10}^{-3}}=2.25\times {{10}^{-4}}{{m}^{3}}$

Energy stored in the capacitor per unit volume is given by,

$u=\frac{{{E}_{1}}}{{{V}_{1}}}=\frac{2.25\times {{10}^{-6}}}{2.25\times {{10}^{-4}}}=0.113J{{m}^{-3}}$

Again, $u=\frac{{{E}_{1}}}{{{V}_{1}}}$

$\frac{V}{d}=$Electric intensity$=E$

$\therefore u=\frac{1}{2}{{\varepsilon }_{0}}{{E}^{2}}$, which is the required energy stored.

18. A $4\mu F$ capacitor is charged by a $200V$ supply. It is then disconnected from the supply, and is connected to another uncharged $2\mu F$ capacitor. How much electrostatic energy of the first capacitor is lost in the form of heat and electromagnetic radiation?

Capacitance of a charged capacitor,  ${{C}_{1}}=4\mu F=4\times {{10}^{-5}}F$

Supply voltage, ${{V}_{1}}=200V$

Electrostatic energy stored in ${{C}_{1}}$ is given by,

${{E}_{1}}=\frac{1}{2}{{C}_{1}}{{V}_{1}}^{2}=\frac{1}{2}\left( 4\times {{10}^{-5}} \right)\times {{\left( 200 \right)}^{2}}=8\times {{10}^{-2}}J$

Capacitance of an uncharged capacitor, ${{C}_{2}}=2\mu F=2\times {{10}^{-6}}F$

When ${{C}_{2}}$ is connected to the circuit, the potential acquired by it is ${{V}_{2}}$,

$\therefore {{V}_{2}}\left( {{C}_{1}}+{{C}_{2}} \right)={{C}_{1}}{{V}_{1}}$

$\Rightarrow {{V}_{2}}\left( 4+2 \right)\times {{10}^{-5}}=4\times {{10}^{-6}}\times 200$

$\Rightarrow {{V}_{2}}=\frac{400}{3}V$

According to the conservation of charge, initial charge on capacitor ${{C}_{1}}$ is equal to the final charge on capacitors ${{C}_{1}}$ and ${{C}_{2}}$.

Electrostatic energy for the combination of two capacitors is given by,

\[{{E}_{2}}=\frac{1}{2}\left( {{C}_{1}}+{{C}_{2}} \right){{V}_{2}}^{1}=\frac{1}{2}\left( 2+4 \right)\times {{10}^{-5}}\times {{\left( \frac{400}{3} \right)}^{2}}=5.33\times {{10}^{-2}}J\]

Hence, the amount of electrostatic energy lost by the capacitor ${{C}_{1}}$ is \[5.33\times {{10}^{-2}}J\]. 

19. Show that the force on each plate of a parallel plate capacitor has a magnitude equal to $\frac{1}{2}QE$, where $Q$ is the charge on the capacitor, and $E$ is the magnitude of the electric field between the plates. Explain the origin of the factor $\frac{1}{2}$.

Ans: Let $F$ be the force applied to separate the plates of a parallel plate capacitor by a distance of$\Delta x$. Hence, work done by the force to do so $=F\Delta x$

As a result, the potential energy of the capacitor increases by an amount given as$uA\Delta x$.

$u=$Energy density

$A=$Area of each plate

$d=$ Distance between the plate

$V=$Potential difference across the plates

The work done will be equal to the increase in the potential energy i.e.,

$F\Delta x=uA\Delta x$

$\Rightarrow F=uA=\left( \frac{1}{2}{{\varepsilon }_{0}}{{E}^{2}} \right)A$

Electric intensity is given by,

$E=\frac{V}{d}$

$\Rightarrow F=\frac{1}{2}{{\varepsilon }_{0}}\left( \frac{V}{d} \right)EA=\frac{1}{2}\left( {{\varepsilon }_{0}}A\frac{V}{d} \right)E$

However, capacitance can be given as, $C=\frac{{{\varepsilon }_{0}}A}{d}$

$\therefore F=\frac{1}{2}\left( CV \right)E$

Charge on the capacitor is given by,

$Q=CV\Rightarrow F=\frac{1}{2}QE$

The physical origin of the factor, $\frac{1}{2}$, in the force formula lies in the fact that just outside the conductor, the field is $E$ and inside it is zero. Hence, it is the average value, $\frac{E}{2}$, of the field that contributes to the force.

20. A spherical capacitor has an inner sphere of radius \[12cm\] and an outer sphere of radius \[13cm\]. The outer sphere is earthed and the inner sphere is given a charge of \[2.5\mu C\]. The space between the concentric spheres is filled with a liquid of dielectric constant \[32\]

(a) Determine the capacitance of the capacitor.

Radius of the inner sphere, \[{{r}_{2}}=12cm=0.12m\]

Radius of the outer sphere, \[{{r}_{1}}=13cm=0.13m\]

Charge on the inner sphere, \[q=2.5\mu C=2.5\times {{10}^{-5}}\]

Dielectric constant of a liquid, \[{{\in }_{r}}=32\]

\[C=\frac{4\pi {{\in }_{0}}{{r}_{1}}{{r}_{2}}}{{{r}_{1}}-{{r}_{2}}}\]

\[{{\in }_{0}}=\]Permittivity of free space \[=8.85\times {{10}^{-12}}{{C}^{2}}{{N}^{-1}}{{m}^{-2}}\]

\[V=\frac{1}{4\pi {{\in }_{0}}}9\times {{10}^{9}}N{{m}^{2}}{{C}^{-2}}\]

\[C=\frac{32\times 0.12\times 0.13}{9\times {{10}^{9}}\times \left( 0.13-0.12 \right)}\Rightarrow C=5.5\times {{10}^{-9}}F\]

Hence, the capacitance of the capacitor is approximately \[5.5\times {{10}^{-9}}F\].

(b) What is the potential of the inner sphere?

Ans: Potential of the inner sphere is given as below

\[\frac{q}{C}=\frac{2.25\times {{10}^{-6}}}{5.5\times {{10}^{-9}}}=4.5\times {{10}^{2}}V\]

Hence, the potential of the inner sphere is \[4.5\times {{10}^{2}}V\].

(c) Compare the capacitance of this capacitor with that of an isolated sphere of radius \[12cm\]. Explain why the latter is much smaller.

Ans:  Given that,

Radius of an isolated sphere given, \[r=12\times {{10}^{-2}}m\]

Capacitance of the sphere is given by the relation,

\[{C}'=4\pi {{\in }_{0}}r\]

\[\Rightarrow {C}'=4\pi \times 8.85\times {{10}^{-12}}\times 12\times {{10}^{-12}}\]

\[\Rightarrow {C}'=1.33\times {{10}^{-11}}F\]

The capacitance of the isolated sphere is less in comparison to the concentric spheres. This is because the outer sphere of the concentric spheres is earthed. Hence, the potential difference is less and the capacitance is more than the isolated sphere.

21. Answer carefully

(a) wo large conducting spheres carrying charges \[{{Q}_{1}}\] and \[{{Q}_{2}}\] are brought close to each other. Is the magnitude of electrostatic force between them exactly given by \[\frac{{{Q}_{1}}{{Q}_{2}}}{4\pi {{\in }_{0}}{{r}^{2}}}\], where \[r\] is the distance between their centres?

Ans: The force between two conducting spheres is not exactly the same given by the expression \[\frac{{{Q}_{1}}{{Q}_{2}}}{4\pi {{\in }_{0}}{{r}^{2}}}\], because there is a non-uniform charge distribution on the sphere.

(b) If Coulomb’s law involved \[\frac{1}{{{r}^{3}}}\] dependence (instead of \[\frac{1}{{{r}^{2}}}\]), would Gauss’s law be still true?

Ans: Gauss’s law will not hold true, if Coulomb’s law involved \[\frac{1}{{{r}^{3}}}\] dependence, instead of \[\frac{1}{{{r}^{2}}}\] on \[r\].

(c) A small test charge is released at rest at a point in an electrostatic field configuration. Will it travel along the field line passing through that point?

Ans: Yes, if a small test charge is released at rest at a point in an electric field configuration, then it will travel along the field lines passing through the point, only if the field lines are straight. This is because the field lines give the direction of acceleration and not of velocity.

(d) What is the work done by the field of a nucleus in a complete circular orbit of the electron? What if the orbit is elliptical?

Ans: Whenever the electron completes an orbit, either circular or elliptical, the work done by the field of a nucleus is zero.

(e) We know that the electric field is discontinuous across the surface of a charged conductor. Is electric potential also discontinuous there?

Ans: No Electric field is discontinuous across the surface of a charged conductor. However, electric potential is continuous.

(f) What meaning would you give to the capacitance of a single conductor?

Ans: The capacitance of a single conductor is considered as a parallel plate capacitor with one of its two plates at infinity.

(g) Guess a possible reason why water has a much greater dielectric constant \[=80\] than say, mica \[=6\].

Ans:   Water has an unsymmetrical space as compared to mica. Since it has a permanent dipole moment, it has a greater dielectric constant than mica.

22. Answer the following

a) The top of the atmosphere is at about \[400kV\] with respect to the surface of the earth, corresponding to an electric field that decreases with altitude. Near the surface of the earth, the field is about \[100V{{m}^{-1}}\]. Why then do we not get an electric shock as we step out of our house into the open? (Assume the house to be a steel cage so there is no field inside).

Ans: We do not get an electric shock as we step out of our house because the original equipotential surfaces of open-air changes, thus maintaining our body and the ground at the same potential.

b) A man fixes outside his house one evening a two-metre-high insulating slab carrying on its top a large aluminium sheet of area \[1{{m}^{2}}\]. Will he get an electric shock if he touches the metal sheet next morning?

Ans: Yes, the man will get an electric shock when he touches the metal slab. The steady discharging current in the atmosphere charges up the aluminium sheet. Resulting in a gradual rise in voltage. The rise in the voltage depends on the capacitance of the capacitor formed by the aluminium slab and the ground.

c) The discharging current in the atmosphere due to small conductivity of air is known to be \[1800A\] on an average over the globe. Why then does the atmosphere not discharge itself completely in due course and become electrically neutral? In other words, what keeps the atmosphere charged?

Ans: The occurrence of thunderstorms and lightning charges the atmosphere continuously. Thus, even with the presence of discharging current of \[1800A\], the atmosphere maintains its electrical neutrality.

d) What are the forms of energy into which the electrical energy of the atmosphere is dissipated during a lightning strike? (Hint: The earth has an electric field of about \[100V{{m}^{-1}}\] at its surface in the downward direction, corresponding to a surface charge density\[=-{{10}^{-9}}C{{m}^{-2}}\]. Due to the slight conductivity of the atmosphere up to about \[50km\] (beyond which is a good conductor), about \[1800C\] is pumped every second into the earth as a whole. The earth, however, does not get discharged since thunderstorms and lightning occurring continually all over the globe pump an equal amount of negative charge on the earth.)

Ans: During lightning and thunderstorms, light energy, heat energy, and sound energy are emitted to the atmosphere.

Class 12 Physics Chapter 2 Important Questions

Chapter 12 - electrostatic potential and capacitance, electrostatic potential .

When a unit positive charge is moved from one point to the other in an electric field, then the amount of work done to move that charge against an electrostatic force with zero acceleration is known as electrostatic potential.

The formula of electrostatic potential is:-

 Electrostatic potential (V) = Work done (W)/Charge(q) 

Volt (V) is the SI unit of electrostatic potential.

When a charge of 1 coulomb is moved is an electric field against an electrostatic force, then the amount of work done is 1J, hence the electrostatic potential will be 1JC-1.

Electrostatic Potential Difference 

When a unit positive charge is moved from say point A to another point say point B against an electrostatic force in an electric field, then the electrostatic potential difference between point A and point B is the amount of work done to move the charge from point A to point B with zero acceleration.

The formula of electrostatic potential difference is:-

              \[ V_{B}-V_{A}=\frac{W_{AB}}{q}\]

Another formula of electrostatic potential difference is:-

              \[ V_{B}-V_{A}=\int_{B}^{A}E.dl\]

The potential difference point A and point B can be found with the line integral of the electric field from point A to point B.

When a point charge q is at any point P and a distance r, then the electrostatic potential is computed:-

              \[V=\frac{1}{4\pi\varepsilon _{o}}.\frac{q}{r}\]

If the point charge is positive, then the electric potential at that point will be positive. Whereas, when the p[oint charge is negative, then the electric potential at that point will be negative.

A positive charge travels from higher potential to lower potential when placed in an electric field. Whereas, negative potential travels from lower potential to higher potential when placed in an electric field.

The Formula of Electrostatic Potential Due to an Electric Dipole

(Image will be uploaded)

The formula of electrostatic potential due to an electric dipole at point P is:-

                \[V=\frac{1}{4\pi\varepsilon _{o}}.\frac{pcos\Theta }{r^{2}}\]

Where, is the angle between p and r.

The Formula of Electrostatic Potential of a Thin Charged Spherical Shell

The formula of the electrostatic potential at a point P inside a thin charged spherical shell that carries charge q and has radius R is:-

\[V=\frac{1}{4\pi\varepsilon _{o}}.\frac{q}{R}\]

The formula of the electrostatic potential at a point P on the surface of a thin charged spherical shell that carries charge q and has radius R is:-

\[V = \frac{1}{4 \pi \epsilon_{0}} . \frac{q}{R}\]

The formula of the electrostatic potential at a point P outside the thin charged spherical shell that carries charge q and the distance between the point P outside the shell and the centre of the shell is ‘r’  is:-

\[V = \frac{1}{4 \pi \epsilon_{0}} . \frac{q}{r}\], where r is greater than R.

Electrostatic Potential Energy

When a charge q is moved from one point to another, the work done by the charge q is stored as electrical potential energy.

The formula of electrostatic potential energy in a system having two charges namely q1 and q2 is:-

\[U = \frac{1}{4 \pi \epsilon_{0}} . \frac{q_{1}q_{2}}{r}\]

These are the important concepts and formulas that are imperative to the students for solving the electrostatic potential and capacitance class 12 important questions. They can also refer to the NCERT question solutions provided by the expert teachers of Vedantu so that grasping of the conceptual part of Physics becomes easy for them.

Important Related Links for CBSE Class 12 Physics 

Mastering the important questions for CBSE Class 12 Physics Chapter 2 - Electrostatic Potential and Capacitance is crucial for success in the upcoming 2024-25 exams. By thoroughly understanding the concepts and solving these questions, students can strengthen their knowledge, improve problem-solving skills, and gain confidence. These questions cover various aspects of electrostatic potential and capacitance, allowing students to deepen their understanding of the topic. With dedicated practice and thorough preparation using these important questions, students can aim for excellent results and excel in their Physics examinations. So, buckle up and embark on this learning journey to achieve academic excellence.

FAQs on Electrostatic Potential and Capacitance Important Questions for CBSE Class 12 Physics Chapter 2

1. How is Electrostatic Potential used for?

Electrostatic potential is used to describe the electric potential energy per unit charge at a point in space. It is a fundamental concept in electrostatics and is used to analyse and understand the behavior of electric fields and charges. Electrostatic potential is used in a wide range of applications, including in the design of electrical devices, the study of biological systems, and the analysis of atmospheric phenomena. It is also used to calculate the work done by electric fields, and to describe the behavior of conductors, dielectrics, and other materials in the presence of electric fields.

2. What is Capacitance used for?

Capacitance is used for storing electrical energy in an electric field. It is commonly used in electronic circuits to store and release electrical energy as needed. Capacitors, which are devices that have capacitance, are used in a wide range of applications, including power supplies, audio filters, timing circuits, and many more. Capacitance also plays a crucial role in many technologies, including telecommunications, computing, and electric vehicles.

3. Can I download the Important Questions for Class 12 Physics Chapter 2 for free?

Yes, you can download the Important Questions for Class 12 Physics Chapter 2 for free from Vedantu. The Important Questions for this chapter are provided in a PDF file on Vedantu. You can refer to these notes online or download them and practice offline. All you need for downloading the Important Questions of Class 12 Physics Chapter 2, Electrostatic Potential and Capacitance, is an internet connection and a digital screen. Also, you can print a hardcopy of these Important Questions for your convenience.

4. What are the uses of Electrostatic Potential and Capacitance according to NCERT for Class 12 Physics Chapter 2?

According to NCERT for Class 12 Physics Chapter 2, the uses of electrostatic potential and capacitance are:

Capacitors: Capacitance is used in capacitors, which are devices used to store electrical energy in an electric field. Capacitors are used in a wide range of applications, including power supplies, audio filters, timing circuits, and many more.

Van de Graaff generator: Electrostatic potential is used in the Van de Graaff generator, which is a device that generates high voltages using electrostatics. This device is used in various scientific experiments, such as particle accelerators, and also in educational demonstrations.

Cathode Ray Oscilloscope: Electrostatic potential is also used in Cathode Ray Oscilloscope (CRO), which is an instrument used to display and analyze electronic signals. The CRO uses electrostatic deflection to move the electron beam across the screen.

Lightning rods: Electrostatic potential is used in lightning rods, which are devices used to protect buildings and structures from lightning strikes. The lightning rod works by creating a path of low resistance for the lightning to follow, thus preventing damage to the structure.

Overall, Electrostatic potential is used in Van de Graaff generators, Cathode Ray Oscilloscopes, and lightning rods. Capacitance is used in capacitors for storing electrical energy in electronic circuits and in a wide range of applications.

5. How many chapters are there in Electrostatic Potential and Capacitance Class 12 Physics?

There is one chapter on Electrostatic Potential and Capacitance in Class 12 Physics. The chapter covers topics such as electric potential, potential difference, capacitance, capacitors, and their applications in electronic circuits and other technologies.

CBSE Class 12 Physics Important Questions

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Class 12th Physics - Electrostatic Potential And Capacitance Case Study Questions and Answers 2022 - 2023

By QB365 on 09 Sep, 2022

QB365 provides a detailed and simple solution for every Possible Case Study Questions in Class 12th Physics Subject - Electrostatic Potential And Capacitance, CBSE. It will help Students to get more practice questions, Students can Practice these question papers in addition to score best marks.

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Electrostatic potential and capacitance case study questions with answer key.

12th Standard CBSE

Final Semester - June 2015

(ii) Which of the following statement is not true?

(iii) Work done in moving a charge from one point to another inside a uniformly charged conducting sphere is

(iv) The work done in bringing a unit positive charge from infinite distance to a point at distance x from a positive charge Q is W. Then the potential  \(\phi\)   at that point is

(v) If  \(1 \mu C\)  charge is shifted from A to B and it is found that work done by an external force is  \(40 \mu \mathrm{J}\) .  In doing so against electrostatics force, the potential difference V A - V B is 

(ii) The change in electric potential energy of the proton for displacement from A to B is

(iii) The mutual electrostatic potential energy between two protons which are at a distance of 9 x 10 -15 m, in  \({ }_{92} \mathrm{U}^{235}\)  nucleus is

(iv) If a system consists of two charges 4 mC and -3mC with no external field placed at (-5 em, 0, 0) and (5 em, 0, 0) respectively. The amount of work required to separate the two charges infinitely away from each other is

(v) As the proton moves from P to Q, then

(ii) The signs of charges Q 1 and Q 2 respectively are

(iii) Which of the two charges Q 1 and Q 2 is greater in magnitude?

(iv) Which of the following statement is not true?

(v) Positive and negative point charges of equal magnitude are kept at  \(\left(0,0, \frac{a}{2}\right)\)  and  \(\left(0,0, \frac{-a}{2}\right)\)  respectively. The work done by the electric field when another positive point charge is moved from (-a, 0, 0) to (0, a, 0) is

(ii) Nature of equipotential surface for a point charge is

(iii) A spherical equipotential surface is not possible

(iv) The work done in carrying a charge q once round a circle of radius a with a charge Q at its centre is

(v) The work done to move a unit charge along an equipotential surface from P to Q

This energy possessed by a system of charges by virtue of their positions. When two like charges lie infinite distance apart, their potential energy is zero because no work has to be done in moving one charge at infinite distance from the other. In carrying a charge q from point A to point B, work done  \(W=q\left(V_{A}-V_{B}\right)\) . This work may appear as change in KE/PE of the charge. The potential energy of two charges q 1 and q 2 at a distance r in air is  \(\frac{q_{1} q_{2}}{4 \pi \varepsilon_{0} r}\) . It is measured in joule. It may be positive, negative or zero depending on the signs of q l and q 2 . (i) Calculate work done in separating two electrons form a distance of 1m to 2m in air, where e is electric charge and k is electrostatic force constant.

(ii) Four equal charges q each are placed at four corners of a square of side a each. Work done in carrying a charge -q from its centre to infinity is

(iii) Two points A and B are located in diametrically opposite directions of a point charge of +2  \(\mu \mathrm{C}\)  at distances 2 m and 1 m respectively from it. The potential difference between A and B is 

(iv) Two point charges A = +3 nC and B = +1 nC are placed 5 ern apart in air. The work done to move charge B towards A by 1 cm is

(v) A charge Q is placed at the origin. The electric potential due to this charge at a given point in space is V. The work done by an external force in bringing another charge q from infinity up to the point is

(ii) How much charge should be placed on a capacitance of 25 pF to raise its potential to l0 5  V?

(iii) Dimensions of capacitance is

(iv) Metallic sphere of radius R is charged to potential V. Then charge q is proportional to

(v) If 64 identical spheres of charge q and capacitance C each are combined to form a large sphere. The charge and capacitance of the large sphere is

(ii) In a parallel plate capacitor, the capacity increases if

(iii) A parallel plate capacitor has two square plates with equal and opposite charges. The surface charge densities on the plates are  \(+\sigma\)  and  \(-\sigma\)   respectively. In the region between the plates the magnitude of the electric field is

(iv) If a parallel plate air capacitor consists of two circular plates of diameter 8 cm. At what distance should the plates be held so as to have the same capacitance as that of sphere of diameter 20 cm?

(v) If a charge of + 2.0 x 10 -8 C is placed on the positive plate and a charge of -1.0 x 10 -8 C on the negative plate of a parallel plate capacitor of capacitance 1.2 x 10 -3   \(\mu \mathrm{F}\)  then the potential difference developed between the plates is

(ii) A parallel plate capacitor with air between the plates has a capacitance of 8 pF. The separation between the plates is now reduced half and the space between them is filled with a medium of dielectric constant 5. Calculate the value of capacitance of the capacitor in second case.

(iii) A dielectric introduced between the plates of a parallel plate condenser

(iv) A parallel plate capacitor of capacitance 1 pF has separation between the plates is d. When the distance of separation becomes 2d and wax of dielectric constant x is inserted in it the capacitance becomes 2 pF. What is the value of x?

(v) A parallel plate capacitor having area A and separated by distance d is filled by copper plate of thickness b. The new capacity is

(ii) A capacitor of capacitance of  \(10 \mu \mathrm{F}\)   is charged to 10 V. The energy stored in it is

(iii) A parallel plate air capacitor has capacity C farad, potential V volt and energy E joule. When the gap between the plates is completely filled with dielectric

(iv) A capacitor with capacitance  \(5 \mu \mathrm{F}\)  is charged to  \(5 \mu \mathrm{C}\) . If the plates are pulled apart to reduce the capacitance to  \(2 \mu \mathrm{F}\) , how much work is done?

(v) A metallic sphere of radius 18 cm has been given a charge of 5 x 10 -6 C. The energy of the charged conductor is

(ii) When air is replaced by a dielectric medium of constant K, the maximum force of attraction between two charges separated by a distance

(iii) Which of the following is a dielectric?

(iv) For a polar molecule, which of the following statements is true?

(v) When a comb rubbed with dry hair attracts pieces of paper. This is because the

Although a single piece of an isolated conductor can store charge on its surface. Ability to store charge is called capacitance (C). If Q is the charge and potential V then \(C=\frac{Q}{V}\)                .. ........(i) If we increase charge, potential on the surface increases, we can store charge only upto some maximum value which is due to some limited maximum potential. Hence, instead of one conducor we use two conductors to form a capacitor, so that more charge can be stored. (i) What is the capacitance of earth. Radius of earth = R = 6.4 x 10 6 m (ii) An uncharged insulated conductor A is brought near a charged insulated conductor B. What happens to charge and potential of B? (iii) What will happen if potential of conductor exceeds its maximum value, so that electric field becomes 3 x 10 6 V/m in air. (iv) On what factors capacitance depends?

Two small charged metal spheres A and Bare situated in a vacuum. The distance between the centres of the sphere is 10 cm. Electric charge on each sphere may be assumed to be a point charge at the centre P of the sphere and is equal to 10 - 9 C each. Point P is a movable point that lies on the line joining the centres of the two charged spheres and is at a distance x from the centre of sphere A. (i) Draw the variation of electric field E with distance x as point P moves towards sphere B. (ii) Also draw the variation of electric potential vs distance x as point P moves towards B. (iii) Calculate electric field V at a distance of 5 cm from centre of sphere A.

A metal sphere of radius 10 cm is charged to a high voltage. If this metal sphere is mounted on a wooden block, then charge will reside on its surface and will not flow from it. As we know dielectric strength of air is 3 x 10 6 V m -1 , then the surrounding air will start conducting and charge stored on the isolated sphere will be lost. (i) Explain why air will start conducting if electric field exceeds dielectric strength? (ii) Calculate the maximum charge this sphere can hold. (iii) Why sometimes electric charge is leaked before the above value of potential is reached.

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Electrostatic potential and capacitance case study questions with answer key answer keys.

(i) (c) (ii) (b) (iii) (a): Since, E = 0 inside the conductor and has no tangential component on the surface, no work is done in moving a small test charge within the conductor and on its surface. (iv) (b) : The work done in bringing unit positive charge from infinity to a point which is at a distance x from the positive charge Q is defined as the potential at the given point due to the charge Q. Therefore \(\phi=W\) (v) (b) :  \(W_{\text {ext }}=q_{0} \Delta V\) \(\left(W_{A B}\right)_{\mathrm{ext}}=q\left(V_{B}-V_{A}\right)\) \(40 \mu \mathrm{J}=1 \mu \mathrm{C}\left(V_{B}-V_{A}\right)\) \(V_{A}-V_{B}=-40 \mathrm{~V}\)

(i) (a) : As  \(\Delta V=-E \Delta \psi=-\left(4.0 \times 10^{8} \mathrm{~V} / \mathrm{m}\right)(0.25 \mathrm{~m})\) = -10 8 V (ii) (c) :  As  \(\Delta U=q_{0} \Delta V=\left(1.6 \times 10^{-19}\right) \times\left(-1.0 \times 10^{8} \mathrm{~V}\right)\) =  \(-1.6 \times 10^{-11} \mathrm{~V}\) (iii) (c) : Here, \(q_{1}=q_{2}=1.6 \times 10^{-19} \mathrm{C}, r=9 \times 10^{-15} \mathrm{~m}\) \(U=\frac{9 \times 10^{9} \times 1.6 \times 10^{-19} \times 1.6 \times 10^{-19}}{9 \times 10^{-15}}=2.56 \times 10^{-14} \mathrm{~J}\) (iv) (a):  Here,  \(q_{1}=4 \mu \mathrm{C}, q_{2}=-3 \mu \mathrm{C}\) r = 10 cm = 0.1 m  Electrostatic potential energy, \(U=\frac{1}{4 \pi \varepsilon_{o}} \frac{q_{1} q_{2}}{r}=9 \times 10^{9} \times \frac{4 \times 10^{-6} \times(-3) \times 10^{-6}}{0.1}=-1.1 \mathrm{~J}\) (v) (a) : As proton moves in the direction of the electric field, then its potential energy decreases.

(i) (c):  \(W=(\text { P.E. })_{\text {final }}-(\text { P.E. })_{\text {initial }}\) \(=\frac{k e^{2}}{2}-\frac{k e^{2}}{1}=\frac{-k e^{2}}{2}\) (ii) (b) : Potential at the centre of the square due to four equal charges q at four corners  \(V=\frac{4 q}{4 \pi \varepsilon_{0}(a \sqrt{2}) / 2}=\frac{\sqrt{2} q}{\pi \varepsilon_{0} a}\) \(W_{0 \rightarrow \infty}=-W_{\infty \rightarrow 0}=-(-q) V=\frac{\sqrt{2} q^{2}}{\pi \varepsilon_{0} a}\) (iii) (c): Here,  \(q=2 \mu \mathrm{C}=2 \times 10^{-6} \mathrm{C}, r_{A}=2 \mathrm{~m}, r_{B}=1 \mathrm{~m}\) \(\therefore \ V_{A}-V_{B}=\frac{q}{4 \pi \varepsilon_{0}}\left[\frac{1}{r_{A}}-\frac{1}{r_{B}}\right]\) \(=2 \times 10^{-6} \times 9 \times 10^{9}\left[\frac{1}{2}-\frac{1}{1}\right] \mathrm{V}=-9 \times 10^{3} \mathrm{~V}\) (iv)(b) : Required work done = Change in potential energy of the system \(W=U_{f}-U_{i}=k \frac{q_{1} q_{2}}{r_{f}}-k \frac{q_{1} q_{2}}{r_{i}}=k q_{1} q_{2}\left[\frac{1}{r_{f}}-\frac{1}{r_{i}}\right]\) \(\therefore \ W=\left(9 \times 10^{9}\right)\left(3 \times 10^{-9} \times 1 \times 10^{-9}\right)\)   \(\times\left[\frac{1}{4 \times 10^{-2}}-\frac{1}{5 \times 10^{-2}}\right]\) \(=27 \times 10^{-7} \times(0.05)=1.35 \times 10^{-7} \mathrm{~J}\) (v) (b)

(i) (b): Here  \(C-50 p F-50 \times 10^{-12} F, V=10^{4} V\) \(R=\frac{1}{4 \pi \varepsilon_{0}} \cdot C=9 \times 10^{9} \mathrm{mF}^{-1} \times 50 \times 10^{-12} \mathrm{~F}\) = 45 x 10 -2 m = 45 cm (ii) (d) : As  \(q=C V=25 \times 10^{-12} \times 10^{5}=2.5 \mu \mathrm{C}\) (iii) (c) (iv) (c): As charge  \(q=C V=\left(4 \pi \varepsilon_{0} R\right) V\) \(\therefore\)   q depends on both V and R. (v) (c): 64 drops have formed a single mop of radius R. Volume oflarge sphere = 64 x Volume of small sphere \(\therefore \frac{4}{3} \pi R^{3}=64 \frac{4}{3} \pi r^{3} \Rightarrow R=4 r \text { and } Q_{\text {total }}=64 q\) \(C^{\prime}=4 \pi \varepsilon_{0} R \Rightarrow C^{\prime}=\left(4 \pi \varepsilon_{0}\right) \cdot 4 r \Rightarrow C^{\prime}=4 C\)

(i) (b): As the capacitor is isolated after charging, charge Q on it remains constant: Plate separation d Increases, capaci.tance decreases as  \(C=\frac{\varepsilon_{0} A}{d}\)  and hence, potential increases as  \(V=\frac{Q}{C}\) . (ii) (c): In a parallel plate capacitor, the capacity of capacitor   \(C=\frac{K \varepsilon_{Q} A}{d} \quad \text { i.e., } C \propto A\) The capacity of capacitor increases if area of the plate increases. (iii) (b): The magnitude of the electric field between the plates is E =  \(\frac{\sigma}{2 \varepsilon_{0}}-\left(-\frac{\sigma}{2 \varepsilon_{0}}\right)=\frac{\sigma}{\varepsilon_{0}}\) (IV) (b): As  \(\frac{\varepsilon_{0} A}{d}=4 \pi \varepsilon_{0} R \text { or } \frac{\varepsilon_{0} \pi D^{2}}{4 d}=4 \pi \varepsilon_{0} R\) \(\text { or } \quad d=\frac{D^{2}}{16 R}=\frac{(0.08)^{2}}{16 \times 0.10}=4 \times 10^{-3} \mathrm{~m}=4 \mathrm{~mm}\) (v) (c): Here  \(V=\frac{q_{1}-q_{2}}{2 C}\) \(=\frac{2.0 \times 10^{-8}+1.0 \times 10^{-8}}{2 \times 1.2 \times 10^{-9}}=12.5 \mathrm{~V}\)

(i) (b):  \(k=\frac{\text { Capacitance with dielectric }}{\text { Capdcitance without dielectric }}\)   \(=\frac{80 \mu \mathrm{F}}{4 \mu \mathrm{F}}=20\) (ii) (c): Capacitance of the capacitor with air between plates \(C^{\prime}=\frac{\varepsilon_{0} A}{d}=8 \mathrm{pF}\) With the capacitor is filled with dielectric (k = 5) between its plates and the distance between the plates is reduced by half, capacitance become  \(C=\frac{\varepsilon_{0} k A}{d / 2}=\frac{\varepsilon_{0} \times 5 \times A}{d / 2}=10 C^{\prime}=10 \times 8=80 \mathrm{pF}\) (iii) (d):  If a dielectric medium of dielectric constant K is filled completely between the plates then capacitance increases by K times . (iv) (b):  \(C=\frac{\varepsilon_{0} A}{d}=1 \mathrm{pF}\)      ....(i) \(C^{\prime}=\frac{x \varepsilon_{0} A}{(2 d)}=2 \mathrm{pF}\)    ....(ii) Divide (ii) by (i), x/2 = 2/1  \(\Rightarrow\)  x = 4 (v) (c) : As capacitance  \(C_{o}=\frac{\varepsilon_{o} A}{d}\) \(\therefore\)  After inserting copper plate,  \(C=\frac{\varepsilon_{o} A}{d-b}\)

(i) (b): As,  \(C_{1}=2 \mu F, C_{2}=4 \mu F\) In series combination, the equivalent capacitance will be,  \(C=\frac{C_{1} C_{2}}{C_{1}+C_{2}}=\left(\frac{2 \times 4}{2+4}\right) \mu \mathrm{F}=\frac{4}{3} \mu \mathrm{F}\) Potential difference applied, V = 6 V Energy stored in the system  \(U=\frac{1}{2} C V^{2}\) \(=\frac{1}{2} \times \frac{4}{3} \times 10^{-6} \times(6)^{2} \mathrm{~J}=24 \mu \mathrm{J}\) (ii) (b) : The energy stored in a capacitor is  \(U=\frac{1}{2} C V^{2}=\frac{1}{2} \times\left(10 \times 10^{-6}\right)(10)^{2}=500 \mu \mathrm{J}\) (iii) (b) : When the gap between the plates is completely filled with dielectric of dielectric constant K, then potential is  \(V=\frac{Q d}{A \varepsilon_{0} K}\)       ... (i) and electric field is \(E=\frac{Q}{A \varepsilon_{0} K}\)    ...(ii) From equations (i) and (ii), both electric field and potential decrease. (iv) (b): Work done =  \(U_{f}-U_{i}=\frac{1}{2} \frac{q^{2}}{C_{f}}-\frac{1}{2} \frac{q^{2}}{C_{i}}\) \(=\frac{q^{2}}{2}\left[\frac{1}{C_{f}}-\frac{1}{C_{i}}\right]=\frac{\left(5 \times 10^{-6}\right)^{2}}{2}\left[\frac{1}{2 \times 10^{-6}}-\frac{1}{5 \times 10^{-6}}\right]\) \(=3.75 \times 10^{-6} \mathrm{~J}\) (v) (b) :  Here  \(r=18 \mathrm{~cm}=18 \times 10^{-2} \mathrm{~m}, q=5 \times 10^{-6} \mathrm{C}\) As  \(C=4 \pi \varepsilon_{0} r=\frac{18 \times 10^{-2}}{9 \times 10^{9}}=2 \times 10^{-11} \mathrm{~F}\) Energy of charged conductor is \(U=\frac{q^{2}}{2 C}=\frac{\left(5 \times 10^{-6}\right)^{2} \mathrm{C}}{2 \times 2 \times 10^{-11} \mathrm{~F} \mid}=0.625 \mathrm{~J}\)

(i) (d): In polar molecule the centres of positive and negative charges are separated even when there is no external field. Such molecule have a permanent dipole moment. Ionic molecule like HCl is an example of polar molecule. (ii) (c): As, \(F_{m}=\frac{F_{o}}{K}\) \(\therefore\)  The maximum force decreases by K times. (iii) (b) (iv) (b) : A polar molecule is one in which the centre of gravity for positive and negative charges are separated. (v) (a)

(i) Given electric field on sphere S 1 E 1 = 2 x 10 4 N/C \(\because \quad E_{1}=2 \times 10^{4} \mathrm{~N} / \mathrm{C}=\frac{k \mathrm{Q}}{r_{1}^{2}}\) Now just inside outer sphere Electric field \( E_{2}=\frac{k \mathrm{Q}}{r_{2}^{2}} \) \(\frac{E_{2}}{E_{1}}=\frac{r_{1}^{2}}{r_{2}^{2}}\) Here r 1 = 0.20 m, r 2 = 0.40 m \(E_{2}=2 \times 10^{4} \times\left(\frac{0.2}{0.4}\right)^{2}=0.5 \times 10^{4} \mathrm{~N} / \mathrm{c}\) (ii) Electrostatic potential inside S 1 \(V_{1}=\frac{k \mathrm{Q}}{r_{1}}=r_{1} E_{1} \quad\left[\because \frac{E_{1}}{V_{1}}=\frac{1}{r_{1}}\right]\) = 0.2 x 2 x 10 4 = 0.4 x 10 4 = 4 x 10 3 V (iii) If S 1 and S 2 are joined by a wire entire amount of energy stored in the system will get converted into heat. Reason: Both the charges on spheres S 1 and S 2 will get neutralized and energy of the system will be dissipated as heat.

(i) Capacitors are used to store electric charges and electric energy. (ii) Both the protons will experience same force. Reason: F = qE; E = constant; q = +e (same) (iii)  \(\because\) V D =.V A > V B = V c \(\therefore\)  Gain in K.E. = q x P.D. \( \therefore\)  Gain in K.E. of proton released from point A will be more. (iv) (a)  \(W_{A \rightarrow B}=e\left(V_{B}-V_{A}\right)\)                    \(\left[\begin{array}{c} \because W=q \times \text { P.D. } \\ E=\frac{V}{d} \end{array}\right]\) = e[E.2d - E.d] = eE.d (b)  \(\because\) V B = V c \(\therefore\)  W BC = 0 (c) W CD = e[V D - V C ] = e[E.(d) - E(2d)] = - eEd (d) W ABCD = W AB + W BC + W CD + W DA = eEd + 0 - eEd + 0 = 0 (v) Electric field is conservative as work done along a closed path is zero.

(i)  \(\mathrm{U}_{\mathrm{A}}=\frac{1}{2} \mathrm{C}_{\mathrm{A}} \mathrm{V}^{2}\) \(=\frac{1}{2} \times\left(2 \times 10^{-6}\right) \times(100)^{2}\) = 10 - 2 J \(\mathrm{U}_{\mathrm{B}}=\frac{1}{2} \mathrm{C}_{\mathrm{B}} \mathrm{V}^{2}\) \(=\frac{1}{2} \times\left(3 \times 10^{-6}\right) \times(100)^{2}\) = 1.5 x 10 - 2 J (ii) Q A = C A V = 2 x 10 -6 x 100 = 2 x 10 -4 C QB = CBV = 3 x 10 -6 x 100 = 3 x 10 -4 C Common Potential \(=\mathrm{V}^{\prime}=\frac{\mathrm{Q}_{\mathrm{B}}-\mathrm{Q}_{\mathrm{A}}}{\mathrm{C}_{\mathrm{A}}+\mathrm{C}_{\mathrm{B}}}=\frac{(3-2) \times 10^{-4}}{(2+3) \times 10^{-6}}\) = 20 V (iii) Charge on capacitor A is Q = CV' = 2 x 10 -6 x 120 = 4 x 10 - 5 C (iv) Here final electrostatic energy \(=\frac{1}{2}\left(\mathrm{C}_{1}+\mathrm{C}_{2}\right)\left(\mathrm{V}^{\prime}\right)^{2}\) \(=\frac{1}{2}(2+3) \times 10^{-6} \times(20)^{2}\) = 10 - 3 J Loss of energy = Initial electrostatic energy - final electrostatic energy = (1.5 + 1.0) x 10 -2 - 10 - 3 = 2.4 x 10 - 2 J

(i) When air molecules/atoms are subjected to high electric field, the centres of negative charges (electrons) and positive charges (nucleus) are separated. Electrons experience force opposite to the direction of electric field and positive charge experiences force in the direction of electric field. If electric field is sufficiently high to pull out electrons, then air gets ionised and starts conducting. (ii)  \(\therefore \ \text { Electric field } \mathrm{E}=\frac{\mathrm{Q}}{4 \pi \varepsilon_{0}}\) and E = 3 x 10 6 V/m then  \(\mathrm{Q}=4 \pi \varepsilon_{0} \times 3 \times 10^{6} \mathrm{C}\) \(\mathrm{Q}=\frac{3 \times 10^{6}}{9 \times 10^{9}}=0.33333 \times 10^{-3} \mathrm{C}\) Q = 333.33 \(\mu\) C (iii) If air is humid, then also it starts conducting and electric charge leaks away.

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• Class 12 Physics MCQs
• Chapter 2 Electrostatic Potential And Capacitance

Class 12 Physics Chapter 2 Electrostatic Potential and Capacitance MCQs

Class 12 Physics Chapter 2 Electrostatic Potential and Capacitance MCQs are provided here with answers. These questions are designed as per the latest CBSE syllabus and NCERT curriculum. Solving these chapter-wise MCQs will help students to score good marks in the final exam. Electrostatic Potential and Capacitance Class 12 Physics MCQs are prepared for a better understanding of the concept. It allows students to test their knowledge and answering skills in the given time frame.

Download Chapter 2 Electrostatic Potential and Capacitance MCQs PDF by clicking on the button below. Download PDF

MCQs on Class 12 Chapter 2 Electrostatic Potential and Capacitance

Check the multiple-choice questions for the 12th Class Physics Electrostatic Potential and Capacitance Chapter. Each MCQ will have four options here, out of which only one is correct. Students have to pick the correct option and check the answer provided here.

1. The process in which a region is made free from any electric field is known as _____________.

• Electrostatic forcing
• Electrostatic binding
• Electrostatic shielding
• None of the options

Answer: (c) Electrostatic shielding

Explanation: Electrostatic shielding is a phenomenon seen when a Faraday cage is used to block the effects of an electric field.

2. The formula for electrostatic potential is _____________.

• Electrostatic potential = Work done*charge
• Electrostatic potential = Work done/charge
• Electrostatic potential = Work done+charge
• Electrostatic potential = Work done-charge

Answer: (b) Electrostatic potential = Work done/charge

Explanation: The formula for electrostatic potential is Electrostatic potential = Work done/charge.

3. 1 Volt = _____________ style-type: lower-alpha;”>

• 1 Newton / 1 Coulomb
• 1 Joule / 1 Coulomb
• 1 Newton / 1 meter

Answer: (c) 1 Joule / 1 Coulomb

Explanation: 1 V = 1 Joule / 1 Coulomb

4. The work done in moving a unit positive test charge over a closed path in an electric field is _____________.

Answer: (c) Zero

Explanation: We say electrostatic forces are conservative in nature since the work done in moving a unit positive test charge over a closed path in an electric field is zero.

5. The electrostatic potential on the perpendicular bisector due to an electric dipole is _____________.

Answer: (a) Zero

Explanation: The electrostatic potential on the perpendicular bisector due to an electric dipole is zero.

6. A surface that has the same electrostatic potential at every point on it is known as _____________.

• Equal-potential surface
• Same potential surface
• Equi-magnitude surface
• Equipotential surface

Answer: (d) Equipotential surface

Explanation: Equipotential surface is a surface formed by the locus of all the points which are at the same potential. Equipotential surfaces do not intersect with each other and are closely spaced in the region of strong electric fields and vice-versa.

7. The work done against electrostatic force gets stored in which form of energy?

• Thermal energy
• Kinetic energy
• Potential energy
• Solar energy

Answer: (c) Potential energy

Explanation: The work done against electrostatic force gets stored in the form of potential energy.

8. Dielectrics are _____________

• Conducting substances
• Non-conducting substances
• Semi-conducting substances
• None of the option

Answer: (b) Non-conducting substances

Explanation: Dielectrics are non-conducting substances.

9. The electric potential inside a conducting sphere _____________

• increases from centre to the surface
• decreases from centre to the surface
• remains constant from centre to the surface

Answer: (d) remains constant from centre to the surface.

Explanation: The electric potential inside a conducting sphere remains constant from centre to the surface.

10. The capacity of the parallel plate capacitor increases when

• area of the plate is decreased
• area of the plate is increased
• distance between the plates increases

Answer: (b) area of the plate is increased

Explanation: The capacity of the parallel plate capacitor increases when the area of the plate is increased.

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Electrostatic Potential And Capacitance MCQ | Class 12 | Physics | Chapter-2

Electrostatic potential and capacitance mcq chapter 2.

Electrostatic potential is the amount of work done to move a charge from a reference point to a specific point.

Below are some of the very important NCERT Electrostatic Potential and Capacitance MCQ Class 12 Physics Chapter 2 with answers. These Electrostatic Potential and Capacitance MCQ have been prepared by expert teachers and subject experts based on the latest syllabus and pattern of CBSE Term 1 examination.

We have given these Electrostatic Potential and Capacitance MCQ Class 12 Physics Questions with Answers to help students understand the concept.

MCQ Questions for Class 12 Physics are very important for the latest CBSE Term 1 and Term 2 pattern. These MCQs are very important for students who want to score high in CBSE Board, NEET and JEE exam. 

We have put together these NCERT MCQ Questions of Electrostatic Potential and Capacitance for Class 12 Physics Chapter 2 with Answers for the practice on a regular basis to score high in exams. Refer to these MCQs Questions with Answers here along with a detailed explanation.

1 . The electric potential inside a conducting sphere

• Increases from centre to surface decreases 
• from centre to surface
• Remains constant from centre to surface 
• Is zero at every point inside

2. One volt is equivalent to

• newton/second
• newton/coulomb
• joule/coulomb   
• joule/second

3. The potential at a point due to a charge of 4 x 10 -7 C located 10 cm away is?

• 3.6 x 10 5 V  
• 3.6 x 10 4 V  
• 4.5 x 10 4 V 
• 4.5 x 10 5 V

4. The electric potential at a point in free space due to a 1 charge Q coulomb is Q x 10 11 V. The electric field at that point is

• 12π? o Q x 10 22 Vm-1 
• 4π? o Q x 10 20 Vm-1
• 12π? o Q x 10 20 Vm-1
• 4π? o Q x 10 22 Vm-1

5. An electric dipole is placed at the centre of a hollow conducting sphere. Which of the following is correct?

• Electric field is zero at every point of the sphere
• Electric field is not zero anywhere on the sphere
• The flux of electric field is not zero through the sphere     
• None of these

6. Figure shows the field lines of a positive point charge. The work done by the field in moving a small positive charge from Q to P is

• zero   
• negative   
• data insufficient

7. The value of electric potential at any point due to any electric dipole is

8. As shown in the figure, charges +q and -q are placed at the vertices B and C of an isosceles triangle. The potential at the vertex A is:

9. Four charges each equal to q are placed at the corners of a square of side l. The electric potential at the centre of the square is

10. Can two equipotential surfaces intersect each other?

• Only when surfaces intersect at 900   

11. If a unit positive charge is taken from one point to  another over an equipotential surface, then

• Work is done on the charge 
• Work is done by the charge  
• Work done is constant
• No work is done

12. A hollow conducting sphere is placed in an electric field produced by a point charge placed at P as shown in figure. Let V A , V B , V C be the potentials at points A, B and C respectively. Then

• V B > V C
• V A > V B

13. A test charge is moved from lower potential point a higher potential point. The potential energy of the charge will

• remains the same    
• decrease   
• becomes zero

14. The potential energy of a system of two charges is negative when

• both the charges are positive
• both the charges are negative
• one charge is positive and other is negative
• both the charges are separated by infinite distance

15. An electric dipole of moment P is placed normal to the lines of force of electric field intensity E, then the work done in deflecting it through an angle of 180° is  

• -2pE   

Click Below To Learn Chemistry Term-1 Syllabus Chapters MCQs

• Chapter-1 : Solid State MCQ
• Chapter-2 : Solution MCQ
• Chapter-7 : P-Block Element MCQ
• Chapter 10 : Haloalkanes and Haloarenes MCQ
• Chapter 11 : Alcohols Phenols and Ether MCQ
• Chapter-14 : Biomolecules MCQ

16. When one electron is taken towards the other electron, then the electric potential energy of system

• decreases   
• remains unchanged 

17. Q amount of electric charge is present on the surface 2 of a sphere having radius R. Then electrical potential energy of this system is  

18. The electrostatic potential energy of a charge of 5C at a point in the electrostatic field is 50 J. The potential at that point is

• 10 v   

19. Two charges of equal magnitude ‘q’ are placed in air at a distance ‘2a’ apart and third charge ‘-2q’ is placed at midpoint. The potential energy of the system is: (? o = permittivity of free space)

• -q 2 /8π? o a
• -3q 2 /8π? o a
• -5q 2 /8π? o a
• -7q 2 /8π? o a

20. A conductor which can be given almost d charge is

• copper   
• gold   

21. If a conductor has a potential zero and there are o charges anywhere else outside, then

• there must be charges on the surface or inside itself
• cannot be any charge in the body of conductor
• there must be charges only on the surface
• both 1 and 2 are correct

22. There are two metallic spheres of same radii but one is solid and the other is hollow, then 

• solid sphere can be given more charge
• hollow sphere can be given more charge 
• they can be charged equally (maximum)
• none of the above

23. A conductor with a positive charge:

• is always at positive potential
• is always at zero potential
• is always at negative potential
• may be at positive, zero or negative potential

24. When air is replaced by a dielectric medium of constant K, the maximum force of attraction between two charges separated by a distance  

• increases K times 
• remains unchanged
• decreases K times   
• increases K-l times

25. Dielectric constant of a medium is also known as

• relative permeability 
• permeability
• permittivity   
• relative permittivity

Click Below To Learn Physics Term 1 Syllabus Chapter-Wise MCQs

• Chapter 1: Electric Charges and Fields MCQ
• Chapter 2: Electrostatic Potential MCQ
• Chapter 3: Current Electricity MCQ
• Chapter 4: Moving Charges And Magnetism MCQ
• Chapter 5: Magnetism And Matter MCQ
• Chapter 6: Electromagnetic Induction MCQ
• Chapter 7: Alternating Current MCQ

26. On decreasing the distance between the plates Of a parallel plate capacitor, its capacitance

• remains unaffected
• first increases then decreases

27. Energy is stored in a capacitor in the form of

• magnetic energy    
• light energy
• heat energy   
• electrostatic energy

28. If in a parallel plate capacitor, which is connected to a battery, we fill dielectrics in whole space of its plates, then which of the following increases?  

• Q and V 
• E and C   

29.  In a parallel plate capacitor, the capacity increases if

• area of the plate is decreased
• distance between the plates increases 
• area of the plate is increased    
• dielectric constant decreases

30. The energy stored in a condenser of capacity C which has been raised to a potential V is given by

• -CV   
• CV   

31. The work done in placing a charge of 8 x 10-18 C on a condenser of capacity 100µF is

• 3.2 x 10 -26 J 
• 3.2 x 10 -31 J 
• 3.2 x 10 -32 J   
• 4 x 10 -26 J

32. A parallel plate capacitor has two square plates with equal and opposite charges. The surface charge  densities on the plates are +σ and -σ respectively. In the region between the plates the magnitude of the electric field is

33. In a charged capacitor, the energy resides

• in the positive charges     
• in both the positive and negative charges
• in the field between the plates
• around the edges of the capacitor plates

34. A parallel plate air capacitor hag capacitance 100 µF. The plates are at a distance d apart. If a slab of thickness t (t < d) and dielectric constant 5 introduced between the parallel plates, then the capacitance will be

35. A parallel plate capacitor is connected across a 2 V battery and charged. The battery is then disconnected and a glass slab is introduced between plates. Which of the following pairs of quantities decreases?

• Charge and potential difference.
• Potential difference and energy stored.
• Energy stored and capacitance.
• Capacitance and charge.

36. An example of an equipotential surface in earth is

A line passing through the centre of the earth connecting two points along the diameter

• A plane that passes through the circular section of the hemisphere of the earth
• A spherical surface at a distance of 1 km from the surface of the earth with its centre at the centre of earth
• a plane on the surface of the earth, which is a tangent to the earth

37. Work done in moving a substance through an equipotential surface is

• depends on the direction of the field

38. Two fixed charges separated by a distance d experience a force F. A dielectric medium of thickness d/4 and dielectric constant 4 is introduces in the space between them. Find the new force acting between the charges.

39. Water is not used as a dielectric between the plates of capacitor because

• dielectric constant is very low
• dielectric strength is very low
• dielectric constant is very high
• dielectric strength is very high

40. 64 small drops of mercury coalesce to form a big drop. The ratio of the surface charge density of each small drop with that of big drop is

41. Electric charge given to the hollow conductor resides

• on the outer surface
• at the centre
• on the inner surface
• uniformly on the outer as well as on the inner surface

42. A man stands inside a large metal charged sphere. Will his hair stand on end?

43. Stainless steel pans are usually provided with copper bottoms. The reason behind this is

• Copper bottoms makes the pan more durable
• Such pans looks colourful
• Copper is easier to clean
• Copper is a better conductor of heat than steel

44. The radii of two metallic spheres are 5 cm and 10 cm and both carry and equal charge of 75µC. If the two spheres are shorted then the charge will be transferred

• 25µC from smaller to bigger
• 25µC from bigger to smaller
• 50µC from smaller to bigger
• 50µC from bigger to smaller

45. A system has been given which has 3 point charges q, 2q and xq which are separated by equal finite distance so that electric potential energy of the system is zero, then the value of x is

Electrostatic Potential and Capacitance MCQ Answers

Electric potential inside a conductor is constant and it is equal to that in the surface of the conductor.

V = q/(4π? o r) = (9 x 10 9 x 4 x 10 -7 ) / 0.1 = 3.6 x 10 4 V

V = Q/(4π? o r) = Q x 1011

4π? o r = 10-11

E = (Q x 4π ? o ) / (4π ? o r) 2 = (Q x 4π ? o ) / (10 -11 ) 2 = 4π ? o Q x 10 22 Vm -1

When electric field is held in the sphere, electric field is not zero anywhere on the sphere. However, net electric flux through the sphere is zero.

In moving a small positive charge from Q to P, work has to be done by an external agency against the electric field. Therefore, work done by the electric field is negative.

V = (pcosΘ) / (4π? o r 2 )

Potential at A = Potential due to +q charge + Potential due to -q charge

q / 4π? o √(a 2 ) – q / 4π? o √(a 2 ) = 0

Electric potential due to each charge at the centre of the square is √2q / 4π? o l

Total potential will be = 4 x (√2q / 4π? o l) = √2q / π? o l

Intersection of two equipotential surfaces at a point will give two directions of electric field intensity at that point, which is not possible.

On the equipotential surface the electric field is normal to the charged surface. So, no work will be done.

Conducting surface behaves as an equipotential surface.

By using U = QV

Q = +1, U = V

At higher potential, potential energy will be high and at lower potential, lower energy will be present

The potential energy is negative whenever there is attraction. Since a positive and negative charge attract each other, their energy is negative. When both the charges are separated by infinite distance, they do not attract each other and their energy is zero .

W = pE (cos 90 o -cos 270 o )

Potential energy of the system U = ((-e) x (-e)) / 4π? o r

Potential = Potential energy / test charge

Earth is conductor which can be given almost unlimited charge.

If a conductor has a non zero potential and there are no charges anywhere else outside, then there must be charges on the surface of the conductor or inside the conductor. There can not be any charge in the body of the conductor.

In case of metallic sphere either solid or hollow, the charge will reside on the surface of the sphere. Since both spheres have same surface area, they can hold amount of maximum charge .

The conductor may be a positive, zero or negative potential. It is according to the way one defines the zero potential .

k = ε r = ε/ε o = Relative permitivity

C = ε o KA / d, capacitance is inversely proportional to the distance.

Energy stored in the capacitor is called electrostatic energy.

C = ε o KA/d and q = CV

Capacitance is directly proportional to the area.

W = q 2 / 2C = ( 8 x 10 -18 ) 2 / 2 x 100 x 10 -6 = 3.2 x 10 -31 J

Energy resides in the field between the plates in a charged capacitor.

Capacitance will increase but not 5 times as the dielectric is not completely filled.

When battery is disconnected, charge remains constant. On introducing glass slab, capacitance increases. Potential difference and energy stored decreases.

Work done in moving a charge over an equipotential surface is always zero.

Water is not used as dielectric between the plates of the capacitor as it has a large dielectric constant but a quite low amount of dielectric strength, which is why it is not used inside a capacitor.

In all conductors, charges reside on the surface.

A metallic sphere means a conductors and as the electric field is zero inside a conductor, the hairs won’t stand up of the man.

Cooper is a better conductor of heat than stainless steel.

Assertion and Reasoning MCQ

(a) Both A and R are true and R is the correct explanation of A

(b) Both A and R are true and but R is not a correct explanation of A

(c) A is true but R is false

(d) A is false, but R is true

1 . Assertion (A) Work done in moving a charge between any two points in an electric field is dependent of the path followed by the charge, between these points.

Reason (R) Electrostatic force is a non-conservative force

2. Assertion (A) The electric potential at any point on the equitorial plane of a dipole is non-zero.

Reason (R) The work done in bringing a unit positive charge from infinity to a point in equitorial plane is not equal for the two charges of the dipole.

3. Assertion (A) For a non-uniformly charged thin circular ring with net charge is zero, the electric field at any point on axis of the ring is zero.

Reason (R) For a non-uniformly charged thin circular ring with net charge zero, the electric potential at each point on axis of the ring is zero.

4. Assertion (A) For a point charge concentric spheres centred at a location of the charge are equipotential surface.

Reason (R) An equipotential surface is a surface over which potential has constant value.

5. Assertion (A) Electric potential and electric potential energy are different quantities

Reason (R) For a system of positive test charge and point charge electric potential energy = electric potential

6. Assertion (A) Surface of a symmetrical conductor can be treated as equipotential surface

Reason (R) Charge can easily flow in a conductor.

7. Assertion (A) Two adjacent conductors of unequal dimensions, carrying the same positive charge have a potential difference between them.

Reason (R) The potential of a conductor depends upon the charge given to it.

8. Assertion (A) When a dielectric slab is gradually inserted between the plates of an isolated parallel plate capacitor, the energy of the system decreases.

Reason (R) The force between the plates decreases.

9. Assertion (A) Polar molecules have temporary dipole moment.

Reason (R) In polar molecule, the centres of positive and negative charges conincide even there is no external field.

10. Assertion (A) In the Van de Graaff generator, the process of spraying the charge is called electron discharge.

Reason (R) Van de Graaff generator produces high voltage and high current.

Assertion-Reason Based Answers

Electrostatic force is a conservative force.

The electric potential at any point on equatorial plane of a dipole is zero.

For a non uniformly charged thin circular ring with net zero charge, electric potential at each point on its axis is zero. Hence electric field at each point on its axis must be perpendicular to the axis.

An equipotential surface is a surface over which potential is same.

Potential and potential energy at different quantities and cannot be equated.

Potential is constant on surface of a sphere so it behaves as an equipotential surface.

Let us considered to spherical shells of radii ‘a’ and ‘b’ which are having the same positive charge Q. Hence, the potential on the surface of each conductor will be,

V1 = kQ/a V2 = kQ/b

As a ≠ b, V1 ≠ V2 and there will be a potential difference too.

C’ = KC and U’ = q 2 /2C’ = q 2 /2KC

The molecules of a substance may be polar or non-polar. In a polar molecule, the centres of positive and negative charges coincide. This molecule has no permanent dipole moment. A polar molecule has its centres of posiitve and negative charges separated, even when there is no electric field. Such molecules have permanent dipole moment.

In Van de Graaff generator, the process of spraying the charge is called corona discharge. Van de Graaff generator produces high voltage and low current.

Case-Study Based MCQ

1. Read the following passage and answer accordingly.

The electric field at a point is the force experienced by unit positive charge at that point. It is a vector quantity. The electric field due to a point charge q at a distance r from it is given by:

Its direction is towards the charge when it is negative and away from the charge when the charge is positive. If there are a system of charges, then field due to each charge will get added vectorially. Electric potential at a point is the work done in bringing unit positive charge from infinity to that point. The potential due to a point charge at a distance r from it is given by

If there are a number of charge in the vicinity of the point, then the potential due to each charge gets add up algebraically (one need to specify the sign of the charge in calculation).

Now consider the following situation.

Four charges +q, +q, -q and -q are placed at the four corners A, B, C and D respectively of a square of side a arranged in the same order. Midpoint of BC is E and that of CD is F. O is the centre of the square.

(i) The direction of the net electric field at O is towards

(a) AB (b) BC (c) CD (d) AD

(ii) The magnitude of the electric field at O is

(iii) The electric potential at O is

(iv) The work done in carrying a change from O to E is

(v) The work done in carrying a charge from O to F is

2. Read the following passage and answer accordingly.

Several capacitors can be connected together to be used in a variety of applications. Multiple connections of capacitors act as a single equivalent capacitor. The total capacitance of this single equivalent capacitor depends both on individual capacitor and how they are connected. Capacitors can be arranged in two simple and common types of connection, known as series and parallel.

(i) 3 capacitors of capacitance 1microF, 2microF and 3microF are connected in series and a potential difference of 11V is applied across the combination. Then potential difference across the plate of 1microF would be\

(a) 2V (b) 4V (c) 1V (d) 6V

(ii) The equivalent capacitance is

(a) 15microF (b) 20microF (c) 25microF (d) 30microF

(iii) The capacitors shown in the following diagram has a capacitance of 4microF each. If C2 capacitor has a capacitance of 10microF, then effective capacitance between A and B is

(a) 2microF (b) 4microF (c) 6microF (d) 8microF

Case-Study Based MCQ Answers

1. (i)(c) (ii)(d) (iii)(d) (iv)(d)

2. (i)(d) (ii)(b) (iii)(b)

Final Words

From the above article, you have practiced Electrostatic Potential and Capacitance MCQ of Class 12 Physics Chapter 2. We hope that the above mentioned latest MCQs for Term 1 of Chapter 2 Electrostatic Potential and Capacitance will surely help you in your exam.

If you have any doubts or queries regarding the Electrostatic Potential and Capacitance MCQ with Answers of CBSE Class 12 Physics, feel free to reach us and we will get back to you as early as possible.

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Class 12 Physics CBSE Electric Potential & Capacitance Board Questions

Here we provide Class 12 Physics important notes,board questions and predicted questions with Answers for chapter Electric Potential & Capacitance. These important notes,board questions and predicted questions are based on CBSE board curriculum and correspond to the most recent Class 12 Physics syllabus. By practising these Class 12 materials, students will be able to quickly review all of the ideas covered in the chapter and prepare for the Class 12 Board examinations.

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Class 12 Physics Assertion Reason Questions Chapter 2 Electrostatic Potential And Capacitance

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Download the chapter on Electrostatic Potential And Capacitance from the CBSE Class 12 Physics Assertion Reason Questions. Class 12 Physics Assertion Reason Questions with Answers were created using the most recent exam format. To gauge their degree of preparedness, students can complete the NCERT Class 12 Physics Chapter 2 Electrostatic Potential And Capacitance.

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Electrostatic Potential And Capacitance Class 12 Assertion Reason Questions

The following questions consist of two statements – Assertion (A) and Reason (R). Answer these questions by selecting the appropriate option given below: (a) Both A and R are true and R is the correct explanation of A. (b) Both A and R are true but R is not the correct explanation of A. (c) A is true but R is false. (d) A is false and R is also false.

Assertion (A) : The electrostatic potential at a point is a scalar quantity. Reason (R) : The electrostatic potential is defined as the work done in bringing a positive test charge from infinity to the point without any acceleration, and work is a scalar quantity.

Answer: (a) Both A and R are true and R is the correct explanation of A.

Assertion (A): Equipotential surfaces are always perpendicular to the electric field lines. Reason (R) : On an equipotential surface, the electric potential is the same at every point, and the electric field is always perpendicular to the surface.

Assertion (A): The potential energy of a charge placed in an electric field is directly proportional to its charge and the electric potential at that point. Reason (R): The potential energy of a charge is given by the equation PE = qV, where q is the charge and V is the electric potential.

Assertion (A) : The electric potential inside a charged conductor is constant and zero at its surface in electrostatic equilibrium. Reason (R): In electrostatic equilibrium, the electric field inside a charged conductor is zero, leading to a constant electric potential throughout the conductor and zero potential at its surface.

Assertion (A): The capacitance of a parallel plate capacitor depends on the area of the plates and the separation between them. Reason (R): The capacitance is directly proportional to the area of the plates and inversely proportional to the separation between them.

Assertion (A): A dielectric material placed between the plates of a capacitor increases its capacitance. Reason (R): A dielectric material reduces the electric field between the plates, resulting in an increase in the capacitance.

Assertion (A) : Capacitors in series have the same charge on each of them. Reason (R): In series, the same charge flows through all the capacitors since they are connected in a closed loop.

Answer: (d) A is false but R is true.

Assertion (A): When a dielectric material is inserted between the plates of a capacitor, the energy stored in the capacitor increases. Reason (R): The presence of a dielectric increases the capacitance, and since energy stored is proportional to the capacitance, the energy stored in the capacitor increases.

Assertion (A): The electric potential inside a charged hollow metallic sphere is zero. Reason (R) : In a hollow charged metallic sphere, the electric field inside is zero, and therefore the electric potential is also zero.

Assertion (A): The capacitance of a capacitor is independent of the charge stored on its plates. Reason (R): The capacitance is determined by the physical characteristics of the capacitor, such as the area of the plates and the distance between them, and is not affected by the amount of charge stored.

Answer: (c) A is true but R is false.

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MCQ Questions for Class 12 Physics Chapter 2 Electrostatic Potential and Capacitance with Answers

We have compiled the NCERT MCQ Questions for Class 12 Physics Chapter 2 Electrostatic Potential and Capacitance with Answers Pdf free download covering the entire syllabus. Practice MCQ Questions for Class 12 Physics with Answers on a daily basis and score well in exams. Refer to the Electrostatic Potential and Capacitance Class 12 MCQs Questions with Answers here along with a detailed explanation.

Electrostatic Potential and Capacitance Class 12 MCQs Questions with Answers

Question 1. Dielectric strength of a medium is 2 KV mm -1 . What is the maximum potential difference that can be set up across a 50 µm specimen without puncturing it. (a) 10 V (b) 100 V (c) 1000 V (d) 10,000 V

Answer: (b) 100 V

Question 2. A charge q is distributed over two spheres of radii r 1 and r 2 such that their surface densities are equal. What is the ratio of their potential. (a) \(\frac {r_1}{r_2}\) (b) \(\frac {r_1^2}{r_2}\) (c) \(\frac {U_1^3}{r_2^3}\) (d) none of these

Answer: (a) \(\frac {r_1}{r_2}\)

Question 3. A hollow metal sphere of radius 5 cm is charged such that the potential on its surface is 10 V. The potential at a distance of 2 cm from the centre of sphere is: (a) Zero (b) 10 V (c) 5 V (d) \(\frac {10}{3}\) V

Answer: (b) 10 V

Question 4. Electric potential V due to an electric dipole is related to the distance r of the observation point as: (a) V ∝ r (b) V ∝ r -1 (c) V ∝ r² (d) V ∝ r -2

Answer: (d) V ∝ r -2

Question 5. Two insulated charged spheres of radii 20 cm and 25 cm respectively and having an equal charge q are connected by a copper wire and then they are separated. (a) Both spheres will have the same charge q. (b) Charge on sphere of radius 20 cm will be more than that of radius 25 cm. (c) Charge on sphere of radius 20 cm will be lesser than that of 25 cm. (d) Charge on each will be zq.

Answer: (c) Charge on sphere of radius 20 cm will be lesser than that of 25 cm.

Question 6. A parallel plate air capacitor is given a charge of 3 µC. A sheet of dielectric constant 3 is inserted so that it completely fills the gap between the plates. The induced charge on each face of the sheet is numericaly: (a) 2 µC (b) 3 µC (c) 0 (d) 9 µC

Answer: (a) 2 µC

Question 7. Three capacitors of capacitances 3µF, 9µF and 18 µF are conected once in series and then in parallel. The ratio of equivalent capacitances C s /C p will be: (a) 1 : 15 (b) 15 : 1 (c) 1 : 1 (d) 1 : 3

Answer: (a) 1 : 15

Question 8. A number of capacitors, each of capacitance 1 µF and each one of which gets punctured if a potential difference just exceeding 500 V is applied are provided. Then an arrangement suitable for giving a capacitor of capacitance 3 µF across which 2000 V may be applied requires at least: (a) 4 component capacitors. (b) 48 component capacitors. (c) 3 component capacitors. (d) 12 component capacitors.

Answer: (b) 48 component capacitors.

Answer: (b) C

Question 10. The capacity of an isolated conducting sphere of radius r is proportional to: (a) r² (b) \(\frac {1}{r}\) (c) \(\frac {1}{r^2}\) (d) r

Answer: (d) r

Question 11. Van-de Graff Generator produces a voltage nearly: (a) 10 6 V (b) 10 8 V (c) 10 4 V (d) none of these.

Answer: (a) 10 6 V

Question 12. The dielectric strength of air at N.T.P. is 3 × 10 6 V/m. Then the maximum charge that can be given to a spherical conductor of radius 3 m is (a) 3 × 10 -1 C (b) 3 × 10² V (c) 3 × 10 -3 C (d) 3 × 10 -4 C

Answer: (c) 3 × 10 -3 C

Question 13. The maximum potential upto which the conductor can be charged in Q 12 is (a) 9 × 10 5 V (b) 9 × 10 6 V (c) 9 × 10 7 V (d) 9 × 10 8 V.

Answer: (b) 9 × 10 6 V

Question 14. ABC is a right angled ∆ with AB = 3 cm, BC = 4 cm, AB = 5 cm. Charges of +15 stat. C, -12 and 20 stat Coulomb are placed at A, B, C respectively. The electric potential energy of three charges is: (a) 180 ergs (b) 4 × 5 × 3 ergs (c) – 60 ergs (d) (20 + 15 – 12) ergs.

Answer: (c) – 60 ergs

Question 15. The work done in moving the three charges to infinite separation is: (a) 180 ergs (b) 60 ergs (c) – 60 ergs (d) (20 + 15 -12) ergs.

Answer: (b) 60 ergs

Fill in the Blanks

Question 1. The charge on the earth cannot be neutralized because ………………..

Answer: Lightning and thunder storms are the continuous phenomenon all over the world which send a charge equal to 1800 C per second into the earth and hence – 5 × 10 5 C charge is maintained on earth.

Question 2. A potential difference of the order or ……………….. is created between the earth and the bottom of the cloud.

Answer: 10 8 V.

Question 3. The top of ……………….. and the surface of the earth form a spherical capacitor of capacitance ………………..

Answer: Stratosphere, 0.09 F (= 0.1 F).

Question 4. Force between the plates of a capacitor is ……………….. which varies ……………….. as the area of the plates and is independent of the ………………..

Answer: \(\frac {1}{2}\), \(\frac {q^2}{ε_0KA}\), inversely, separation between the plates.

Question 5. When a dielectric slab is introduced by disconnecting the battery from the capacitor, then q = ……………….. V = ……………….. σ = ……………….. E = ……………….. U = ……………….. C = ………………..

Answer: q = q 0 σ = σ 0 E = \(\frac {V_0}{K}\) U = \(\frac {U_0}{K}\) C = KC 0 .

Question 6. If the medium surrounding the conductor is charged, then the potential of the conductor is ………………..

Answer: Changed.

Question 7. If at any point E = 0, then V may or may not be ……………….. e.g. inside a charged conductor E = ……………….. but V is ………………..

Answer: Zero, Zero, not zero.

Question 8. Electric potential at a point on the equitorial line of a dipole is ……………….. but \(\vec{E}\) is ………………..

Answer: Zero, not equal to zero.

Question 9. Two equipotential surfaces cannot intersect each other because ………………..

Answer: Then there will be two values of electric potential at that point which is impossible.

Question 10. Energy of an electric dipole is minimum when ………………..

Answer: It is aligned with the direction of the electric field.

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Electrostatic Potential & Capacitance Class 12 Important Questions with Solutions

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Physics topic Electrostatic Potential & Capacitance is an important lesson for students to master because various questions are asked from it. However, for assistance in CBSE Class 12 Physics board exam preparation here on this page, we have shared the direct link of Electrostatic Potential & Capacitance Class 12 Important Questions with Solutions.

Along with the PDF link, we have also shared how the important questions on Electrostatic Potential & Capacitance play an important role.

Why Solve Electrostatic Potential & Capacitance Class 12 Important Questions?

Due to various reasons, Class 12 students should be thorough with Electrostatic Potential & Capacitance Class 12 important questions. A few reasons to solve CBSE Class 12 Physics Electrostatic Potential & Capacitance important questions are -

• To Enahce Grip on Electrostatic Potential & Capacitance: To develop a deeper-level understanding of Electrostatic Potential & Capacitance students should solve important questions on Class 12 Physics Electrostatic Potential & Capacitance. The important questions are generally considered a good level of question that enable students to use all of their acquired Physics knowledge to solve the questions.
• To Practise Numerous Board-Level Physics Questions: There is nothing better than solving actual board-level questions and therefore, CBSE Class 12 students should solve the PDF file of Electrostatic Potential & Capacitance Class 12 Important Questions with Solutions because solving the PDF file will help students to not only practise numerous board-level Physics questions but to get instant answers too.
• For Boosting Confidence: Being able to solve actual board-level questions give students confidence, at the same time students refer to the Electrostatic Potential & Capacitance Class 12 Important Questions with Solutions are able to boost their confidence. Overall, we can say that solving the important Class 12 Physics questions using the answers PDF helps students to level up their board exam preparation and gives them the confidence to answer questions in the actual board examination.
• To Prepare for Class 12 Electrostatic Potential & Capacitance Important Questions: The Electrostatic Potential & Capacitance important questions are a collection of those questions which have a higher possibility to be asked in the upcoming CBSE Class 12 Physics examination. However, students solving the important questions on this topic will be able to better prepare for the Class 12 Physics exam.

Features of CBSE Class 12 Physics Electrostatic Potential & Capacitance Class 12 Important Questions PDF

In this section, we are discussing the features of Class 12 Physics Important Questions PDF that will help students like you to better understand why the PDF provided here is the best to use.

• All Types of Questions: One mark to 5 marks all types of questions are mentioned in the PDF file of Electrostatic Potential & Capacitance Class 12 Important Questions with Solutions.
• Physics Electrostatic Potential & Capacitance Questions with Answers: The subject experts have answered the Physics Electrostatic Potential & Capacitance important questions so, the solutions are 100% accurate and can be used by students to solve their unsolved questions.
• Diagrams are Used: Physics has lots of diagrams, images and graphics that play a crucial role in understanding the concepts and questions. Therefore, the subject matter experts have included the images and diagram in Electrostatic Potential & Capacitance Class 12 Important Questions with Solutions PDF.
• Prepared by Physics Experts: To maintain the accuracy and authenticity of Class 12 Electrostatic Potential & Capacitance Important Questions and its solutions PDF, the subject experts have prepared the PDF file’s content.

How to Download Class 12 Electrostatic Potential & Capacitance Important Questions with Answers?

Here’s a step-by-step process to download the Class 12 Electrostatic Potential & Capacitance Important Questions with Answers PDF. 

• Go to Selfstudys.com using an internet browser
• After reaching the homepage of Selfstudys.com click on the navigation icon/button

• A side pop-up bar will appear, navigate CBSE and click on that

• It will bring more options to choose from, click on the previous year question paper

• A new page will load, click on “12th PYP Chapter Wise”

• Scroll the page a little and click on Physics then find the PDF file of Electrostatic Potential & Capacitance Class 12 Important Questions with Solutions.

How to Use Important Questions for Class 12 Physics Electrostatic Potential & Capacitance PDF?

One can use important questions for Class 12 Physics Electrostatic Potential & Capacitance PDF in whatever way they want, however, there are certain ways that we think are a good way to utilise the important questions for Class 12 Physics Electrostatic Potential & Capacitance PDF.

• To Revise Important Portions of Electrostatic Potential & Capacitance: The PDF file of Class 12 Physics Electrostatic Potential & Capacitance can be a great study tool to revise the concepts and subtopics discussed in this lesson. Furthermore, the students can also use the same PDF file to recall the questions based on the Electrostatic Potential & Capacitance.
• To Do Last-Minute Exam Preparation: Students preparing for the CBSE Class 12 Board exam and are in the last-minute exam preparation process can use the PDF file of Electrostatic Potential & Capacitance important questions undoubtedly.
• Find Weakness to Improve: The Electrostatic Potential & Capacitance important question class 12 PDF can be a great study resource to find weak areas in this lesson. After finding those weaker areas students can use the answers provided to improve the errors and to learn something new. Students can also ask their elders, teachers or tutors to help them with the Electrostatic Potential & Capacitance important questions to master them.

When to Use Class 12 Electrostatic Potential & Capacitance Important Questions with Solutions PDF?

Once a Class 12 student has studied their Electrostatic Potential & Capacitance lesson completely can go for the important questions to practise. However, there are a few more times when a CBSE Class 12 student can utilise the PDF file of Electrostatic Potential & Capacitance Class 12 Important Questions with Solutions 

• When Doing Board Exam Preparation: One can refer to the Class 12 Physics Electrostatic Potential & Capacitance important questions when they are doing the board exam preparation. At the time of preparation, referring to the PDF file is very helpful because it helps students better understand the types of questions that are asked in the CBSE Class 12 Physics Board examination. Also, solving the Class 12 Electrostatic Potential & Capacitance important questions during board exam preparation helps students refresh their answering methods for a variety of questions that can be asked from the chapter Electrostatic Potential & Capacitance.
• While Solving Chapter-End Questions: One of the best times to use the PDF file of CBSE Class 12 Physics Electrostatic Potential & Capacitance is when solving the Chapter-end questions. It is the best time because at this time students have a very fresh knowledge of Electrostatic Potential & Capacitance and therefore, they would be able to answer most of the important questions. However, if one struggles to answer the important questions given in the PDF file, one would be required to read the same chapter again to strengthen their grip.

Precautions to Take When Using Electrostatic Potential & Capacitance Class 12 Important Questions with Solutions PDF

There are certain points that students should take care of when using Electrostatic Potential & Capacitance Class 12 Important Questions PDF. Those points are mentioned below:

• Read the Questions Carefully: It is essential to read the questions carefully to better understand the Electrostatic Potential & Capacitance Class 12 Important Questions and to answer them. Reading the questions carefully, ease the process of answering questions.
• Answer Questions Based on the Exam Pattern: The PDF file of CBSE Class 12 Physics Electrostatic Potential & Capacitance important questions contains details on whether a particular question is one word or long answer types of questions. So, it is advised students answer Electrostatic Potential & Capacitance Class 12 Important Questions following the exam pattern or as directed in the PDF file.
• Don’t Completely Rely on Electrostatic Potential & Capacitance Class 12 Important Questions with Solutions PDF: No doubt, Electrostatic Potential & Capacitance important question Class 12 is a great study tool and it can help students better prepare for the board examination of Physics, however, students should make sure not to be dependent on the PDF file of Electrostatic Potential & Capacitance important question class 12.

How Electrostatic Potential & Capacitance Class 12 Important Questions with Solutions PDF are Prepared?

Various study materials such as previous year question papers, sample question papers, CBSE Class 12 Physics Syllabus and several other study resources are used to prepare the Electrostatic Potential & Capacitance Class 12 Important Questions with Solutions PDF.

• Using Previous Year Question Papers of Class 12 Physics: One of the ideal study materials that are used to prepare the CBSE Class 12 Physics Electrostatic Potential & Capacitance Important Questions are the sets of Previous Year Question Papers. It helps subject experts to find and match the most frequent and high-scoring Electrostatic Potential & Capacitance questions to add them to the PDF file.
• Taking Help of Class 12 Physics Syllabus: The Class 12 Physics Syllabus also plays an important role because it helps students better predict the important questions of Electrostatic Potential & Capacitance. It is possible because the syllabus contains a complete detail of the Electrostatic Potential & Capacitance weightage for the board examination.

How to Deal With CBSE Class 12 Electrostatic Potential & Capacitance Important but Difficult Questions?

While solving the CBSE Class 12 Electrostatic Potential & Capacitance important questions if a student feels stuck, they can use the following methods to deal with them.

• Use the Answers Provided in PDF: The PDF file of Electrostatic Potential & Capacitance Class 12 Important Questions with Solutions contains very accurate answers prepared by subject experts. The answers are easier to understand too so when feeling stuck at some Electrostatic Potential & Capacitance questions refer to the answers available in the PDF.
• Consult Teachers/Tutors: If provided answers don’t help much to the students then they should consult teachers or tutors for help. They will guide them in understanding the questions and answering them with ease.
• Use the Internet: If above mentioned things don’t help much the Internet will surely help students to understand and answers the important questions of Class 12 Electrostatic Potential & Capacitance. There are several resources on the internet that can help students better prepare for the exam.

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