NCERT Solutions for Class 12 Physics Chapter 2 Electrostatic Potential and Capacitance

Class 12 Physics Chapter 2 Electrostatic Potential and Capacitance is one of the important study materials. This chapter includes usage of many complicated equations and formulas that students learn in their Class 12. A PDF file is available for free download which includes important questions, answers to questions from textbook, worksheets, and other assignments where a student can apply the formulas learnt during the chapter to solve these questions.

Download NCERT Solutions Class 12 Physics Chapter 2 PDF:-Download Here

NCERT solutions for class 12 is a complete package that will help students score well in their examinations.

It is essential to score good marks in class 12th board examination because the marks scored in class 12 board examination will help you to take admission in your preferred college. NCERT solutions Physics Chapter 2 Electrostatic Potential and Capacitance is provided here so that students can refer to these NCERT solutions for better understanding and clarification of the chapter.

Class 12 Physics NCERT solutions for Electrostatic Potential and Capacitance

This chapter provides good marks weightage to derivations and numerical problems. The derivation of topics like potential energy of the system of charges, potential due to electric dipole and energy stored in the capacitor is frequently asked in exams. Numerical problems to find the effective capacitance of a complex combination of capacitors is asked in exams. A few questions regarding tracing the equipotential surface for dipole or system of charges are asked in exams. All the topics in the chapter are covered in the NCERT Solutions Class 12 Physics Chapter 2.

Subtopics of class 12 NCERT Physics Electrostatic Potential and Capacitance

Section Number Topic
2.1 Introduction
2.2 Electrostatic Potential
2.3 Potential Due To A Point Charge
2.4 Potential Due To An Electric Dipole
2.5 Potential Due To A System Of Charges
2.6 Equipotential Surfaces
2.6.1 Relation Between Field And Potential
2.7 Potential Energy Of A System Of Charges
2.8 Potential Energy In An External Field
2.8.1 Potential Energy Of A Single Charge
2.8.2 Potential Energy Of A System Of Two Charges In An External Field
2.8.3 Potential Energy Of A Dipole In An External Field
2.9 Electrostatics Of Conductors
2.10 Dielectrics And Polarization
2.11 Capacitors And Capacitance
2.12 The Parallel Plate Capacitor
2.13 Effect Of Dielectric On Capacitance
2.14 Combination Of Capacitors
2.14.1 Capacitors In Series
2.14.2 Capacitors In Parallel
2.15 Energy Stored In A Capacitor
2.16 Van De Graaff Generator

 

Class 12 Physics NCERT Solutions Electrostatic Potential and Capacitance Important Questions


Q 2.1) Two charges 5 × 10-8 C and -3 x 10-8 C are the two charges located 16 cm apart from each other. At what point(s) on the line joining the two charges is the electric potential zero? Take the potential at infinity to be zero.

Soln.: Given,

q1 = 5 x 10-8 C

q2 = -3 x 10-8 C

The two charges are at a distance, d = 16cm = 0.16m from each other.

As shown in the figure, let us consider a point P over the line joining charges q1 and q2.

Let, distance of the considered point P from q1 be ‘r’

Let, point P has zero electric potential (V).

The electric potential at point P is the summation of potentials due to charges q1 and q2.

Therefore,  V=14πϵo.q1r+14πϵo.q2dr V = \frac{1}{4\pi \epsilon _{o}} . \frac{q_{1}}{r} + \frac{1}{4\pi \epsilon _{o}}.\frac{q_{2}}{d – r}  ……….(1)

Here,

ϵo \epsilon _{o} = permittivity of free space.

Putting  V = 0, in eqn. (1), we get,

0 = 14πϵo.q1r+14πϵo.q2dr \frac{1}{4\pi \epsilon _{o}} . \frac{q_{1}}{r} + \frac{1}{4\pi \epsilon _{o}}.\frac{q_{2}}{d – r} 14πϵo.q1r=14πϵo.q2dr\frac{1}{4\pi \epsilon _{o}}.\frac{q_{1}}{r} = -\frac{1}{4\pi \epsilon _{o}}.\frac{q_{2}}{d – r} q1r=q2dr\frac{q_{1}}{r} = – \frac{q_{2}}{d – r} 5×108r=(3×108)0.16r\frac{5 \times 10^{-8}}{r} = – \frac{(- 3 \times 10^{-8})}{0.16 – r}

5(0.16 – r)  = 3r

0.8 = 8r

r = 0.1m = 10 cm.

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

Let us assume a point P at a distance ‘s’  from the negative charge be outside the system, having potential  zero.

So, for the above condition, potential is given by –

V=14πϵo.q1s+14πϵo.q2sd V = \frac{1}{4\pi \epsilon _{o}} . \frac{q_{1}}{s} + \frac{1}{4\pi \epsilon _{o}}.\frac{q_{2}}{s – d} ………………………(2)

Here,

ϵo \epsilon _{o} = permittivity of free space.

For V = 0, eqn. (2) can be written as :

0 = 14πϵo.q1s+14πϵo.q2sd \frac{1}{4\pi \epsilon _{o}} . \frac{q_{1}}{s} + \frac{1}{4\pi \epsilon _{o}}.\frac{q_{2}}{s – d} 14πϵo.q1s=14πϵo.q2sd \frac{1}{4\pi \epsilon _{o}}.\frac{q_{1}}{s} = -\frac{1}{4\pi \epsilon _{o}}.\frac{q_{2}}{s – d} q1s=q2sd \frac{q_{1}}{s} = – \frac{q_{2}}{s – d} 5×108s=(3×108)s0.16 \frac{5 \times 10^{-8}}{s} = – \frac{(- 3 \times 10^{-8})}{s – 0.16}

5(s – 0.16) = 3s

0.8  = 2s

S = 0.4 m = 40 cm.

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

 

Q 2.2)  A regular hexagon of side 10 cm has a charge 5 µC at each of its vertices. Calculate the potential at the centre of the hexagon.

Soln.:

The figure shows, regular hexagon containing charges q, at each of its vertices.

Here,

q =  5 µC = 5 x 10-6 C.

Length of each side of hexagon, AB =BC = CD = DE = EF = FA = 10 cm.

Distance of the vertices from the centre O, d = 10 cm.

Electric potential at point O,

V =  14πϵo.6xqd \frac{1}{4\pi \epsilon _{o}}.\frac{6 x q}{d}

Here,

ϵo \epsilon _{o} = Permittivity of free space and 14πϵo \frac{1}{4\pi \epsilon _{o}} = 9 x 109 Nm2C-2

V =  9x109 x6x5x1060.1 \frac{9 x 10^{9} x 6 x 5 x 10_{-6}}{0.1} = 2.7 x 106 V.

 

Q 2.3) Two charges 2 µC and -2 µC are placed at points A and B 6 cm apart.

(1) Identify an equipotential surface of the system.

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

Soln.:

(1) An equipotential surface is defined as the surface over which the total potential is zero. In the given question this plane is normal to line AB. The plane is located at the mid – point of the line AB as the magnitude of the charges are same.

 

(2) At every point on this surface the direction of the electric field is normal to the plane in the direction of AB.

 

Q 2.4) A spherical conductor of radius 12 cm has a charge of 1.6 x 10-7C distributed uniformly on its surface. What is the electric field

(1) inside the sphere.

(2) just outside the sphere.

(3) at a point 18 cm from the centre of sphere.

Soln.:

(1) Given,

Radius of spherical conductor, r = 12cm = 0.12m

Charge is distributed uniformly over the surface, q = 1.6 x 10-7 C.

Electric field inside a spherical conductor is zero.

 

(2) Electric field E, just outside the conductor is given by the relation,

E =  14πϵo.qr2 \frac{1}{4\pi \epsilon _{o}}.\frac{q}{r^{2}}

Here,

= permittivity of free space and 14πϵo \frac{1}{4\pi \epsilon _{o}} = 9 x 109 Nm2C-2

Therefore,

E = 9×109×1.6×107(0.12)2 \frac{9 \times 10^{9} \times 1.6 \times 10^{-7}}{(0.12)^{2}}

Therefore, just outside the sphere the electric field is 4.4 x 104 NC-1.

 

(3) From the centre of sphere the electric field at a point 18m = E1.

From the centre of sphere the distance of point d = 18 cm = 0.18m.

E1 = 14πϵo.qd2 \frac{1}{4\pi \epsilon _{o}} . \frac{q}{d^{2}} = 9×109×1.6×107((1.8×10)2)2 \frac{9 \times 10^{9} \times 1.6 \times 10^{-7}}{((1.8 \times 10)^{-2})^{2}} = 4.4 x 104 NC-1

So, from the centre of sphere the electric field at a point 18 cm away is 4.4 x 104 NC-1.

 

Q 2.5) A parallel plate capacitor with air between the plates has a capacitance of 8pF (1pF = 10-12 F. 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?

Soln.: Given,

Capacitance, C = 8pF.

In first case the parallel plates are at a distance ‘d’  and is filled with air.

Air has dielectric constant, k = 1

Capacitance, C = k×ϵo×Ad \frac{k \times \epsilon _{o} \times A}{d} = ϵo×Ad \frac{\epsilon _{o} \times A}{d}      … eq(1)

Here,

A = area of each plate

ϵo \epsilon _{o} = permittivity of free space.

Now, if the distance between the parallel plates is reduced to half, then d1 = d/2

Given, dielectric constant of the substance, k1 = 6

Hence, the capacitance of the capacitor,

C1 = k1×ϵo×Ad1 \frac{k_{1} \times \epsilon _{o} \times A}{d_{1}} = 6ϵo×Ad/2 \frac{6 \epsilon _{o} \times A}{d/2} = 12ϵoAd \frac{12 \epsilon _{o} A}{d}   … (2)

Taking ratios of eqns. (1) and (2), we get,

C1 = 2 x 6 C = 12 C = 12 x  8 pF = 96pF.

Hence, capacitance between the plates is 96pF.

 

Q 2.6)  Three capacitors connected in series have capacitance of 9pF each.

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

(2) What is the potential difference across each capacitor if the combination is connected to a 120 V supply?

Soln.:

(1)  Given,

The capacitance of the three capacitors, C = 9 pF

Equivalent capacitance (ceq) is the capacitance of the combination of the capacitors given by

1Ceq=1C+1C+1C=3C=39=13 \frac{1}{C_{eq}} = \frac{1}{C} + \frac{1}{C} + \frac{1}{C} = \frac{3}{C} = \frac{3}{9} =\frac{1}{3} 1Ceq=13 \frac{1}{C_{eq}} = \frac{1}{3} = Ceq = 3 pF

Therefore, the total capacitance = 3pF.

 

(2) Given, supply voltage, V = 100v

The potential difference (V1) across the capacitors will be equal to one – third of the supply voltage.

Therefore, V1V3 \frac{V}{3} = 1203 \frac{120}{3} = 40V.

Hence, the potential difference across each capacitor is 40V.

 

Q 2.7)  Three capacitors of capacitances 2 pF, 3 pF and 4 pF are connected in parallel.

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

(2) Determine the charge on each capacitor if the combination is connected to a 100 V supply.

Soln.:

(1) Given, C1 = 2pF, C2 = 3pF and C3 = 4pF.

Equivalent capacitance for the parallel combination is given by Ceq .

Therefore, Ceq = C1 + C2 + C3 = 2 + 3 + 4 = 9pF

Hence, total capacitance of the combination is 9pF.

(2) supply voltage, V = 100v

The three capacitors are having the same voltage, V = 100v

q = vc

where,

q = charge

c = capacitance of the capacitor

v = potential difference

for capacitance, c = 2pF

q = 100 x 2 = 200pC = 2 x 10-10C

for capacitance, c = 3pF

q = 100 x 3 = 300pC = 3 x 10-10C

for capacitance, c = 4pF

q = 100 x 4 = 400pC = 4 x 10-10 C

 

Q 2.8) In a parallel plate capacitor with air between the plates, each plate has an area of 6 x 10-3 m 2 and and the distance between the plates is 3 mm. Calculate the capacitance of the capacitor. If this capacitor is connected to a 100 V supply, what is the charge on each plate of the
capacitor?

Soln.:  Given,

The area of plate of the capacitor, A = 6 x 10-3 m2

Distances between the plates, d = 3mm = 3 x 10-3 m

Voltage supplied, V = 100v

Capacitance of a parallel plate capacitor is given by, C =  ϵ×Ad \frac{\epsilon \times A}{d}

Here,

ε = permittivity of free space = 8.854 × 10-12 N-1 m -2 C-2

C = 8.854×1012×6×1033×103\frac{8.854 \times 10^{-12} \times 6 \times 10^{-3}}{3 \times 10^{-3}} = 17.81 x 10-12 F = 17.71 pF.

Therefore, each plate of the capacitor is having a charge of

q = VC = 100 × 17.81 x 10-12 C = 1.771 x 10-9 C

 

Q 2.9: Explain what would happen if, in the capacitor given in Exercise 2.8, a 3 mm thick mica sheet (of dielectric constant = 6) were inserted between the plates,

(a)while the voltage supply remained connected.

(b)after the supply was disconnected.

Answer 2.9:

(a) Dielectric constant of the mica sheet, k = 6

If voltage supply remained connected, voltage between two plates will be

constant.

Supply voltage, V = 100 V

Initial capacitance, C = 1.771 × 10−11 F

New capacitance, C1 = kC = 6 × 1.771 × 10−11 F = 106 pF

New charge, q1 = C1V = 106 × 100 pC = 1.06 × 10–8 C

Potential across the plates remains 100 V.

(b) Dielectric constant, k = 6

Initial capacitance, C = 1.771 × 10−11 F

New capacitance, C1 = kC = 6 × 1.771 × 10−11 F = 106 pF

If supply voltage is removed, then there will be constant amount of charge

in the plates.

Charge = 1.771 × 10−9 C

Potential across the plates is given by,

V1 = q/C1 = 1.771×109106×1012\frac{1.771 \times 10^{-9}}{106 \times 10^{-12}}

= 16.7 V

 

Q 2.10) A 12pF capacitor is connected to a 50V battery. How much electrostatic energy is stored in the capacitor?

Soln.: Given,

Capacitance of the capacitor, C = 12pF = 12 x 10-12 F

Potential difference, V = 50 V

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

E = 12 \frac{1}{2} CV2 = 12 \frac{1}{2} x 12 x 10-12 x (50)2 J = 1.5 x 10-8 J

Therefore, the electrostatic energy stored in the capacitor is 1.5 x 10-8 J.

was disconnected.

 

Q 2.11) A 600pF capacitor is charged by a 200V supply. It is then disconnected from the supply and is connected to another uncharged 600 pF capacitor. How much electrostatic energy is lost in the process?

Soln.: Given,

Capacitance, C = 600pF

Potential difference, V = 200v

Electrostatic energy stored in the capacitor is given by :

E1 = 12 \frac{1}{2}CV2 = 12 \frac{1}{2} x (600 x 10-12) x (200)2J = 1.2 x 10-5 J

Acc. to the question, the source is disconnected to the 600pF and connected to another capacitor of 600pF, then equivalent capacitance (Ceq) of the combination is given by,

1Ceq \frac{1}{C_{eq}} = 1C \frac{1}{C} + 1C \frac{1}{C} 1Ceq \frac{1}{C_{eq}} = 1600 \frac{1}{600} + 1600 \frac{1}{600}

= 2600 \frac{2}{600} = 1300 \frac{1}{300}

Ceq = 300pF

New electrostatic energy can be calculated by:

E2 = 12 \frac{1}{2}CV2 = 12 \frac{1}{2} x 300 x (200)2 J = 0.6 x 10-5 J

Loss in electrostatic energy,

E = E1 – E2

E = 1.2 x 10-5 – 0.6 x 10-5 J = 0.6 x 10-5 J = 6 x 10-6 J

Therefore, the electrostatic energy lost in the process is 6 x 10-6 J.

 

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