NCERT Solutions For Class 12 Physics Chapter 2

NCERT Solutions Class 12 Physics Electrostatic Capacitance

Ncert Solutions For Class 12 Physics Chapter 2 PDF Free Download

NCERT solutions for class 12 physics Electrostatic Capacitance is an important study material. In class 12 physics, there are many complicated equations and formulas which are used to solve questions related to electrostatic capacitance. 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 for class 12 physics chapter 2 is provided here so that students can refer to these ncert solutions for better understanding and clarification of the chapter. The class 12 Physics Chapter 2 includes the following topics

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 Polarisation
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

Q.1)  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) between these two charges the electric potential is 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,  \( V = \frac{1}{4\pi \epsilon _{o}} . \frac{q_{1}}{r} + \frac{1}{4\pi \epsilon _{o}}.\frac{q_{2}}{d – r} \)  ……….(1)

Here,

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

Putting  V = 0, in eqn. (1), we get,

0 = \( \frac{1}{4\pi \epsilon _{o}} . \frac{q_{1}}{r} + \frac{1}{4\pi \epsilon _{o}}.\frac{q_{2}}{d – r} \)

\(\frac{1}{4\pi \epsilon _{o}}.\frac{q_{1}}{r} = -\frac{1}{4\pi \epsilon _{o}}.\frac{q_{2}}{d – r}\)

\(\frac{q_{1}}{r} = – \frac{q_{2}}{d – r}\)

\(\frac{5 \times 10^{-8}}{r} = – \frac{(- 3 \times 10^{-8})}{0.16 – r}\)

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 –

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

Here,

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

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

0 = \( \frac{1}{4\pi \epsilon _{o}} . \frac{q_{1}}{s} + \frac{1}{4\pi \epsilon _{o}}.\frac{q_{2}}{s – d} \)

\( \frac{1}{4\pi \epsilon _{o}}.\frac{q_{1}}{s} = -\frac{1}{4\pi \epsilon _{o}}.\frac{q_{2}}{s – d} \)

\( \frac{q_{1}}{s} = – \frac{q_{2}}{s – d} \)

\( \frac{5 \times 10^{-8}}{s} = – \frac{(- 3 \times 10^{-8})}{s – 0.16} \)

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)  A regular hexagon having sides equal to 10 cm, contains charges of 5 µC at its each vertices.

Find the potential at the centre of this regular 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 =  \( \frac{1}{4\pi \epsilon _{o}}.\frac{6 x q}{d} \)

Here,

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

V =  \( \frac{9 x 10^{9} x 6 x 5 x 10_{-6}}{0.1} \) = 2.7 x 106 V.

 

Q.3)  Charges 2 µC and -2 µC at a point A and B, respectively are 6 cm apart.

(1) Obtain an equipotential surface of the system.

(2) At every point on the surface find the direction of electric field.

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.4) A conductor is in the shape of a sphere with radius 12 cm having charge 1.6 x 10-7C uniformly distributed over its surface. What will be 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 =  \( \frac{1}{4\pi \epsilon _{o}}.\frac{q}{r^{2}} \)

Here,

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

Therefore,

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

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 = \( \frac{1}{4\pi \epsilon _{o}} . \frac{q}{d^{2}} \) = \( \frac{9 \times 10^{9} \times 1.6 \times 10^{-7}}{((1.8 \times 10)^{-2})^{2}} \) = 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.5) What will be the capacitance of a parallel plate capacitor with air between the plates  having capacitance of 8pF (1pF = 10-12 F). How much the capacitance of the capacitor will change  if the distance between the plate is reduced to half, and the space between the plates 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 = \( \frac{k \times \epsilon _{o} \times A}{d} \) = \( \frac{\epsilon _{o} \times A}{d} \)     … eq(1)

Here,

A = area of each plate

\( \epsilon _{o} \) = 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 = \( \frac{k_{1} \times \epsilon _{o} \times A}{d_{1}} \) = \( \frac{6 \epsilon _{o} \times A}{d/2} \) = \( \frac{12 \epsilon _{o} A}{d} \)  … (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.6)  Three capacitors connected in series have capacitance of 9pF each.

(1)  What can be total capacitance of this combination?

(2) Find the potential difference across each capacitor if they are connected to a 120v supply?

Soln.:

(1)  Given,

Capacitance of the three capacitors, C = 9 pF

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

\( \frac{1}{C_{eq}} = \frac{1}{C} + \frac{1}{C} + \frac{1}{C} = \frac{3}{C} = \frac{3}{9} =\frac{1}{3} \)

\( \frac{1}{C_{eq}} = \frac{1}{3} \) = Ceq = 3 pF

Therefore, 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, V1 =  \( \frac{V}{3} \) = \( \frac{120}{3} \) = 40V.

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

 

Q.7)  In parallel combination three capacitors of capacitances 2pF, 3pF and 4pF are connected.

(1) What will be the total capacitance of this combination?

(2) If the connection is connected to 100V supply, find the charge on each capacitor.

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.8) A parallel plate capacitor having air in between the plates each having area of 6 x 10-3 m 2 and distance between them is 3 mm. Find the capacitance of the capacitor. What will be the charge on each plate if the capacitor is connected to a 100v source?

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 =  \( \frac{\epsilon \times A}{d} \)

Here,

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

C = \(\frac{8.854 \times 10^{-12} \times 6 \times 10^{-3}}{3 \times 10^{-3}}\) = 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 9 :Considering the same capacitor as in Q 8, when a thick mica sheet, 3mm wide is placed between the plates (Dielectric constant = 6 )

(a)With constant voltage supply

(b)  After disconnecting the voltage supply

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 = \(\frac{1.771 \times 10^{-9}}{106 \times 10^{-12}}\)

= 16.7 V

 

Q.10) How much of electrostatic energy is stored in the capacitor if the capacitor with 12pF of capacitance is connected to a 50v battery?

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 = \( \frac{1}{2}\) CV2 = \( \frac{1}{2}\) 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.11) 200V of source is connected to a 600pF capacitor. Later, the source is disconnected to the 600pf capacitor and connected to another 600pF capacitor. How much of 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 = \( \frac{1}{2}\)CV2 = \( \frac{1}{2}\) 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,

\( \frac{1}{C_{eq}} \) = \( \frac{1}{C} \) + \( \frac{1}{C} \)

\( \frac{1}{C_{eq}} \) = \( \frac{1}{600} \) + \( \frac{1}{600} \)

= \( \frac{2}{600} \) = \( \frac{1}{300} \)

Ceq = 300pF

New electrostatic energy can be calculated by:

E2 = \( \frac{1}{2}\)CV2 = \( \frac{1}{2}\) 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.

This chapter provides a good weightage to derivations and numericals. The derivation of topics like potential energy of system of charges, potential due to electric dipole and energy stored in the capacitors is frequently asked in exams. Numericals to find the effective capacitance of 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 12. Download the free PDF provided here, if necessary, take a printout to keep it handy during the preparation of exams.

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