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

NCERT Solutions for Class 12 Physics Chapter 2 – Free PDF Download

NCERT Solutions for Class 12 Physics Chapter 2 Electrostatic Potential and Capacitance includes the usage of many complicated equations and formulas that students learn in their Class 12. A PDF file of the NCERT Solutions for Class 12 Physics Electrostatic Potential and Capacitance is available here for free download. The PDF includes important questions, answers to questions from the textbook, worksheets and other assignments.

These NCERT Solutions for Class 12 is essential to score good marks in the Class 12 term – I examination. These solutions are curated by individual subject matter experts according to the latest update on the term – I CBSE Syllabus (2021-22) and its guidelines. Further, the NCERT Solutions for Class 12 Physics Chapter 2 Electrostatic Potential and Capacitance provided here can be used by students to understand the concepts discussed in the chapter in detail.

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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 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.

Solution: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.

Let us consider a point “P” over the line joining charges q1 and q2.

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

Let us consider point P to have 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, the 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.

Solution:

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

Figure

Here,

q =  5 µC = 5 × 10-6 C.

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

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

The electric potential at point O,

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

Here,

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

V =  9×109 ×6×5×1060.1 \frac{9 × 10^{9} × 6 × 5 × 10^{-6}}{0.1} = 2.7 × 106 V.

 

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

(1) Identify the equipotential surface of the system.

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

Solution:

(1) An equipotential surface is defined as the surface over which the total potential is zero. In the given question the 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.

Solution:

(1) Given,

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

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

The 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,

ε0 = 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=105NC1 \frac{9 \times 10^{9} \times 1.6 \times 10^{-7}}{(0.12)^{2}}=10^5\,NC^{-1}

Therefore, just outside the sphere the electric field is 105 NC-1.

 

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

From the centre of the 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(1.8×102)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?

Solution: Given,

Capacitance, C = 8pF.

In the 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, the 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?

Solution:

(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.

Solution:

(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, the 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 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?

Solution:  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,

εo = 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}} = 17.71 x 10-12 F = 17.71 pF.

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

q = VC = 100 × 17.71 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, the 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 remain 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 the supply voltage is removed, then there will be a 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?

Solution: 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.

 

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?

Solution: 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

According 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 10-12 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.

Q2.12) A charge of 8 mC is located at the origin. Calculate the work done in taking a small charge of –2 × 10–9 C from a point P (0, 0, 3 cm) to a point Q (0, 4 cm, 0), via a point R (0, 6 cm, 9 cm).

NCERT Class 12 - Chapter 2 - Image - Question 2.12

Solution:

Charge located at the origin, q = 8 mC = 8 x 10-3 C

The magnitude of the charge taken from the point P to R and then to Q, q1 = – 2 x 10-9 C

Here OP= d1= 3 cm = 3 x 10-2 m

OQ = d2= 4 cm = 4 x 10-2 m

Potential at the point P, V1=q4πϵ0d1V_{1}=\frac{q}{4\pi \epsilon _{0}d_{1}}

Potential at the point Q, V2=q4πϵ0d2V_{2}=\frac{q}{4\pi \epsilon _{0}d_{2}}

The work done (W) is independent of the path

Therefore, W = q1[V1 – V2]

 

W=q1[q4πϵ0d2q4πϵ0d1]W=q_{1}\left [ \frac{q}{4\pi \epsilon _{0}d_{2}} – \frac{q}{4\pi \epsilon _{0}d_{1}} \right ]

 

W=qq14πϵ0[1d21d1]W=\frac{qq_{1}}{4\pi \epsilon _{0}}\left [ \frac{1}{d_{2}} – \frac{1}{d_{1}} \right ]

Where, 14πϵ0=9×109Nm2C2\frac{1}{4\pi \epsilon _{0}}=9 \times 10^{9}Nm^{2}C^{-2}

Therefore,

W=9×109×8×103×(2×109)[14×10213×102]W = 9 \times 10^{9}\times 8\times 10^{-3}\times (-2\times 10^{-9})\left [ \frac{1}{4\times 10^{-2}}-\frac{1}{3\times 10^{-2}} \right ]

= -144 x 10-3 x (-100/12)

= 1.2  Joule

Therefore, the work done during the process is 1.2 J

Q2.13) 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.

Solution:

NCERT Class 12 - Chapter 2 - Image - Question 2.13

Sides of the cube = b

Charge at the vertices = q

Diagonal of one of the sides of the cube

d2=b2+b2=2b2d^{2}= \sqrt{b^{2}+b^{2}}= \sqrt{2b^{2}}

d = b√2

Length of the diagonal of the cube

l2=d2+b2=2b2+b2=3b2l^{2}= \sqrt{d^{2}+b^{2}}= \sqrt{2b^{2}+b^{2}}=\sqrt{3b^{2}}

l = b√3

The distance between one of the vertices and the centre of the cube is

r = l/2 =(b√3/2)

The electric potential (V) at the centre of the cube is due to the eight charges at the vertices

V = 8q/4πε0

V=8q4πϵ0(b32)V = \frac{8q}{4\pi \epsilon _{0}\left ( b\frac{\sqrt{3}}{2} \right )} =4q3πϵ0b= \frac{4q}{\sqrt{3}\pi \epsilon _{0}b}

Therefore, potential at the centre of the cube is 4q3πϵ0b\frac{4q}{\sqrt{3}\pi \epsilon _{0}b}

The electric field intensity at the centre of the cube, due to the eight charges is zero.  The charges are distributed symmetrically with respect to the centre of the cube. Therefore, they get cancelled.

Q2.14) Two tiny spheres carrying charges 1.5 µC and 2.5 µC are located 30 cm apart. Find the potential and electric field:
(a) at the mid-point of the line joining the two charges, and
(b) at a point 10 cm from this midpoint in a plane normal to the
line and passing through the mid-point.

NCERT Class 12 - Chapter 2 - Image - Question 2.14

Solution:

Two tiny spheres carrying charges are located at points A and B

The charge at point A, q1= 1.5 µC

The charge at point B, q2 = 2.5 µC

The distance between the two charges = 30 cm  = 0.3 m

(a) Let O be the midpoint. Let V1 and E1 be the potential and electric field respectively at the midpoint.

V1 = Potential due to charge at A + Potential due to charge at B

V1=q14πϵ0(d2)+q24πϵ0(d2)V_{1}=\frac{q_{1}}{4\pi \epsilon _{0}(\frac{d}{2})}+ \frac{q_{2}}{4\pi \epsilon _{0}(\frac{d}{2})} V1=14πϵ0(d2)(q1+q2)V_{1}=\frac{1}{4\pi \epsilon _{0}(\frac{d}{2})}(q_{1}+q_{2})

Here,

ε0 = Permittivity of the free space

14πϵ0=9×109NC2m2\frac{1}{4\pi \epsilon _{0}} = 9 \times 10^{9}NC^{2}m^{-2}

Therefore,

V1=9×109×106(0.302)(2.5+1.5)=2.4×105VV_{1}=\frac{9\times 10^{9}\times 10^{-6}}{(\frac{0.30}{2})}(2.5+ 1.5)= 2.4\times 10^{5}V

Electric field at O, E1 = Electric field due to q2 – Electric field due to q1

=q24πϵ0(d2)2q14πϵ0(d2)2=\frac{q_{2}}{4\pi \epsilon _{0}(\frac{d}{2})^{2}}- \frac{q_{1}}{4\pi \epsilon _{0}(\frac{d}{2})^{2}} =9×109(0.302)2×106×(2.51.5)=\frac{9\times 10^{9}}{(\frac{0.30}{2})^{2}}\times 10^{-6}\times (2.5-1.5)

= 400 x 103 V/m

Therefore, the potential at midpoint is 2.4 x 105 V and the electric field at the midpoint is 400 x 103 V/m.

(b) Consider a point Z such that the distance OZ = 10 cm = 0.1 m as shown in the figure.

Let V2 and E2 be the potential and electric field respectively at point Z.  The distance

BZ=AZ=(0.1)2+(0.15)2=0.18mBZ = AZ = \sqrt{(0.1)^{2}+(0.15)^{2}}= 0.18 m

The potential at V2 = Potential due to the charge at A + Potential due to the charge at B

=q14πϵ0(AZ)+q24πϵ0(BZ)= \frac{q_{1}}{4\pi \epsilon _{0}(AZ)}+\frac{q_{2}}{4\pi \epsilon _{0}(BZ)}

 

=9×109×1060.18(1.5+2.5)=\frac{9\times 10^{9}\times 10^{-6}}{0.18}(1.5+2.5)

= 2 x 105 V

The electric field due to q1 at Z

EA=q14πϵ0(AZ)2E_{A}=\frac{q_{1}}{4\pi \epsilon _{0}(AZ)^{2}} =9×109×1.5×106(0.18)2= \frac{9\times 10^{9}\times 1.5\times 10^{-6}}{(0.18)^{2}}

= 416 x 103 V/m

The electric field due to q2 at Z

EB=q24πϵ0(BZ)2E_{B}=\frac{q_{2}}{4\pi \epsilon _{0}(BZ)^{2}} =9×109×2.5×106(0.18)2= \frac{9\times 10^{9}\times 2.5\times 10^{-6}}{(0.18)^{2}}

= 694 x 103 V/m

The resultant field intensity at Z

E=EA2+EB2+2EAEBcos2θE=\sqrt{E_{A}^{2}+E_{B}^{2}+2E_{A}E_{B}cos2\theta }

From the figure we get cos θ = (0.10/0.18) = 5/9 = 0.5556

θ = cos -1 (0.5556) = 56.25

2θ = 2 x 56.25 = 112.50

cos 2θ = – 0.38

E=(0.416×106)2+(0.69×106)2+2×0.416×0.69×1012×(0.38)E=\sqrt{(0.416\times 10^{6})^{2}+(0.69\times 10^{6})^{2}+2\times 0.416\times 0.69\times 10^{12}\times (-0.38)}

= 6.6 x 105 V/m

Therefore the potential at the point Z is 694 x 103 V/m and the electric field is 6.6 x 105 V/m.

Q2.15) A spherical conducting shell of inner radius r1 and outer radius r2 has a charge Q.
(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?
(b) Is the electric field inside a cavity (with no charge) zero, even if the shell is not spherical, but has any irregular shape? Explain.

Solution:

(a) If a charge +q is placed at the centre of the shell, a charge of magnitude -q is induced in the inner surface of the shell. Therefore, the surface charge density at the inner surface of the shell is given by the relation,

σ1 = Total charge / Inner surface area = – q/4πr12——–(1)

A charge +q is induced on the outer surface of the shell. The total charge on the outer surface of the shell is Q+q. Surface charge density at the outer surface of the shell

σ2 = Total charge / Outer surface area = (Q+q)/4πr12————(2)

(b) Yes. The electric field inside a cavity (with no charge) will be zero, even if the shell is not spherical but has any irregular shape. Take a closed-loop, part of which is inside the cavity along a field line and the rest inside the conductor. Since the field inside the conductor is zero, this gives a net work done by the field in carrying a test charge over a closed loop. We know this is impossible for an electrostatic field. Hence, there are no field lines inside the cavity (i.e., no field), and no charge on the inner surface of the conductor, whatever be its shape.

Q2.16) (a) Show that the normal component of electrostatic field has a discontinuity from one side of a charged surface to another given by
(E2E1).n^=σϵ0\left ( E_{2}-E_{1} \right ).\hat{n}=\frac{\sigma }{\epsilon _{0}}
where n^\hat{n} is a unit vector normal to the surface at a point and σ is the surface charge density at that point. (The direction of  n^\hat{n} is from side 1 to side 2.) Hence, show that just outside a conductor, the electric field is σn^/ϵ0\sigma \hat{n}/\epsilon _{0}

(b) Show that the tangential component of the electrostatic field is continuous from one side of a charged surface to another. [Hint: For (a), use Gauss’s law. For, (b) use the fact that work done by electrostatic field on a closed loop is zero.].

Solution:

(a) Let E1 be the electric field at one side of the charged body and E2 is the electric field at the other side of the charged body. If the infinite plane charged body has a uniform thickness, the electric field due to one of the surface of the charged body is

E1=σ2ϵ0n^\vec{E_{1}}=-\frac{\sigma }{2\epsilon _{0}}\hat{n} ——–(1)

here,

n^\hat{n} = unit vector normal to the surface at a point

σ = surface charge density at that point

The electric field due to the other surface of the charged body is

E2=σ2ϵ0n^\vec{E_{2}}=\frac{\sigma }{2\epsilon _{0}}\hat{n} ——–(2)

The electric field at any point due to the charge surfaces

E2E1=σ2ϵ0n^+σ2ϵ0n^=σϵ0n^\vec{E_{2}}-\vec{E_{1}}=\frac{\sigma }{2\epsilon _{0}}\hat{n} + \frac{\sigma }{2\epsilon _{0}}\hat{n} = \frac{\sigma }{\epsilon _{0}}\hat{n}

Since inside the conductor, E1=0\vec{E_{1}}= 0

 

E2E1=σϵ0n^\vec{E_{2}}-\vec{E_{1}}= \frac{\sigma }{\epsilon _{0}}\hat{n} ——(3)

Therefore, the electric field just outside the conductor is σϵ0n^\frac{\sigma }{\epsilon _{0}}\hat{n}

(b)  When a charged 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 the electrostatic field is continuous from one side of a charged surface to the other.

Q2.17) A long charged cylinder of linear charged density λ is surrounded by a hollow co-axial conducting cylinder. What is the electric field in the space between the two cylinders?

Solution:

Let the length of the charged cylinder and the hollow co-axial conducting cylinder be L

Charge density of the long charged cylinder is λ

Let E be the electric field in the space between the two cylinders

According to Gauss theorem, the electric flux through the Gaussian surface is given as Φ = E (2πd)L

Here,

d is the distance between the common axis of the cylinders

Therefore , Φ = E (2πd)L = q/ε0

here, q is the charge on the inner surface of the outer cylinder

ε0  is the permittivity of the free space

E (2πd)L = λL/ε0

E = λ/2πdε0

Therefore, the electric field in the space between the two cylinders is λ/2πdε0

Q2.18) In a hydrogen atom, the electron and proton are bound at a distance of about 0.53 Å:
(a) Estimate the potential energy of the system in eV, taking the zero of the potential energy at infinite separation of the electron from proton.
(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)?
(c) What are the answers to (a) and (b) above if the zero of potential
energy is taken at 1.06 Å separation?

Solution:

The distance between the proton and electron of hydrogen atom, d = 0.53 Å

Charge of the electron, q1 = -1.6 x 10-19 C

Charge of the proton,  q2 = +1.6 x 10-19 C

(a) At infinite separation of electron and proton, potential energy is zero

Potential energy of the system = Potential energy at infinity – Potential energy at distance d

= 0 – q1q24πϵ0d\frac{q_{1}q_{2}}{4\pi \epsilon _{0}d}

here,

ε0 is the permittivity of the free space

14πϵ0=9×109Nm2C2\frac{1}{4\pi \epsilon _{0}}=9×10^{9}Nm^{2}C^{-2}

Potential energy= 0 – 9×109×(1.6×1019)20.53×1010=43.7x1019J\frac{9\times 10^{9}\times (1.6 \times 10^{-19})^{2}}{0.53\times 10^{10}} = -43.7 x 10^{-19}J

Potential energy = -43.7 x 10-19/1.6 x 10-19 = -27.2 eV [Since 1.6 x 10-19 J = 1 eV]

Therefore, the potential energy of the system is -27.2 eV

(b) Half of the magnitude of the potential energy is equal to the kinetic energy

Kinetic energy = |V|/2 = (1/2) x (27.2) = 13. 6 eV

Total energy = Kinetic energy + potential energy

= 13.6 eV – 27.2 eV

Total energy = – 13.6 eV

Therefore, the minimum work required to free an electron is – 13.6 eV

(c) When the zero of the potential energy is taken as, d1= 1.06 Å

Potential energy of the system = Potential energy at d1 – Potential energy at d

q1q24πϵ0d1\frac{q_{1}q_{2}}{4\pi \epsilon _{0}d_{1}} – 27.2 eV

9×109×(1.6×1019)21.06×1010\frac{9\times 10^{9}\times (1.6 \times 10^{-19})^{2}}{1.06\times 10^{10}} -27.2 eV

= 21.73 x 10-19 J – 27.2 eV

= 13.58 eV -27.2 eV   [Since 1.6 x 10-19 J = 1 eV]

= -13.6 eV

Q2.19) If one of the two electrons of the H2 molecule is removed, we get a hydrogen molecular ion H2+. In the ground state of an H2+, the two protons are separated by roughly 1.5 Å, and the electron is roughly 1 Å from each proton. Determine the potential energy of the system. Specify your choice of the zero of potential energy.

Solution:

Charge of the 1st proton, q1 = 1.6 x 10-19 C

Charge of the 2nd proton, q2 = 1.6 x 10-19 C

Charge of the electron, q3 = -1.6 x 10-19 C

Distance between the 1st and the 2nd proton, d1 = 1.5 x 10-10 m

Distance between the 1st proton and the electron, d2 = 1 x 10-10 m

Distance between the 2nd proton and the electron, d3 = 1 x 10-10 m

The potential energy at infinity is zero

Therefore, the potential energy of the system is

V=q1q24πϵ0d1+q2q34πϵ0d3+q3q14πϵ0d2V= \frac{q_{1}q_{2}}{4\pi \epsilon _{0}d_{1}}+\frac{q_{2}q_{3}}{4\pi \epsilon _{0}d_{3}}+ \frac{q_{3}q_{1}}{4\pi \epsilon _{0}d_{2}}

 

V=14πϵ0[q1q2d1+q2q3d3+q3q1d2]V = \frac{1}{4\pi \epsilon _{0}}\left [ \frac{q_{1}q_{2}}{d_{1}}+\frac{q_{2}q_{3}}{d_{3}}+\frac{q_{3}q_{1}}{d_{2}}\right ]

Substituting (1/4πε0) = 9 x 109 Nm2C-2 we get

V=9×109[1.6×1019×1.6×10191.5×1010+1.6×1019×(1.6×1019)1×1010+(1.6×1019)(1.6×1019)1×1010]V= 9\times 10^{9}\left [ \frac{1.6\times 10^{-19}\times1.6\times 10^{-19} }{1.5\times 10^{-10}}+\frac{1.6\times 10^{-19}\times (-1.6\times 10^{-19})}{1\times 10^{-10}}+\frac{(-1.6\times 10^{-19})(1.6\times 10^{-19})}{1\times 10^{-10}}\right ]

 

V=9×109×1019×10191010[(1.6)21.5(1.6)2(1.6)2]V=\frac{9\times 10^{9}\times 10^{-19}\times 10^{-19}}{10^{-10}}\left [ \frac{(1.6)^{2}}{1.5}-(1.6)^{2}-(1.6)^{2} \right ]

=-30.7 x 10-19 J

=-19.2 eV (1eV = 1.6 x 10-19 J)

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

Q2.20) 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 the charge density on the sharp and pointed ends of a conductor is higher than on its flatter portions.

Solution:

Let A be the sphere of radius a, Charge QA and capacitance CA

Let B be the sphere of radius b, Charge QB and capacitance CB

The conducting spheres are connected by a wire, therefore the potential of both the capacitors will be V

The electric field due to a, EA=QA4πϵ0a2E_{A}=\frac{Q_{A}}{4\pi \epsilon _{0}a^{2}}

The electric field due to b, EB=QB4πϵ0b2E_{B}=\frac{Q_{B}}{4\pi \epsilon _{0}b^{2}}

The ratio of electric fields at the surface of the spheres is

EAEB=QA4πϵ0a2×b24πϵ0QB\frac{E_{A}}{E_{B}}=\frac{Q_{A}}{4\pi \epsilon _{0}a^{2}}\times \frac{b^{2}4\pi \epsilon _{0}}{Q_{B}}

 

EAEB=QAQB×b2a2\frac{E_{A}}{E_{B}}=\frac{Q_{A}}{Q_{B}}\times \frac{b^{2}}{a^{2}} ——-(1)

 

QAQB=CAVCBV\frac{Q_{A}}{Q_{B}}=\frac{C_{A}V}{C_{B}V}

 

QAQB=ab\frac{Q_{A}}{Q_{B}}=\frac{a}{b}——–(2)

Putting equation (2) in equation (1) we get

EAEB=ab×b2a2\frac{E_{A}}{E_{B}}=\frac{a}{b}\times \frac{b^{2}}{a^{2}}

Therefore, the ratio of the electric field at the surface is b/a

A sharp and pointed end is like a sphere of very small radius and the flat portion is like a sphere of large radius. Therefore, the charge density is of pointed ends is higher than the flat portion.

Q2.21) Two charges –q and +q are located at points (0, 0, –a) and (0, 0, a), respectively.
(a) What is the electrostatic potential at the points (0, 0, z) and (x, y, 0) ?
(b) Obtain the dependence of potential on the distance r of a point from the origin when r/a >> 1.
(c) How much work is done in moving a small test charge from the point (5,0,0) to (–7,0,0) 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?

Solution:

(a)  Two charges –q and +q are located at points (0, 0, –a) and (0, 0, a), respectively. They will form a dipole. The point (0, 0, z) is on the axis of the dipole and (x,y,0) is normal to the dipole. The electrostatic potential at (x,y,0) is zero. The electrostatic potential at (0,0,z) is given by

V=14πϵ0(qza)+14πϵ0(qz+a)V = \frac{1}{4\pi \epsilon _{0}}\left ( \frac{q}{z-a} \right )+\frac{1}{4\pi \epsilon _{0}}\left ( -\frac{q}{z+a} \right )

 

V=q(z+az+a)4πϵ0(z2a2)V = \frac{q(z+a-z+a)}{4\pi \epsilon _{0}(z^{2}-a^{2})}

 

V=q(2a)4πϵ0(z2a2)V = \frac{q(2a)}{4\pi \epsilon _{0}(z^{2}-a^{2})} =p4πϵ0(z2a2)= \frac{p}{4\pi \epsilon _{0}(z^{2}-a^{2})}

ε0 = Permittivity of free space

p = dipole moment of the system= q x 2a

(b) The distance “r” is much larger than half of the distance between the two charges. Therefore, the potential at the point r is inversely proportional to the square of the distance.i.e., V∝(1/r2).

(c) x,y plane is a equipotential surface and x-axis is a equipotential line. Therefore, the change in potential (dV) along x-axis will be zero. The work done in moving a small test charge from the point (5,0,0) to (–7,0,0) along the x-axis is given by

Potential at (5,0,0)

V1=14πϵ0(q(50)2a2)+14πϵ0(q(50)2(a)2)=0V_{1} = \frac{1}{4\pi \epsilon _{0}}\left ( \frac{q}{\sqrt{(5-0)^{2}-a^{2}}} \right )+\frac{1}{4\pi \epsilon _{0}}\left ( -\frac{q}{\sqrt{(5-0)^{2}-(-a)^{2}}} \right )=0

Potential at (-7,0,0)

V2=14πϵ0(q((7)0)2a2)+14πϵ0(q((7)0)2(a)2)=0V_{2} = \frac{1}{4\pi \epsilon _{0}}\left ( \frac{q}{\sqrt{((-7)-0)^{2}-a^{2}}} \right )+\frac{1}{4\pi \epsilon _{0}}\left ( -\frac{q}{\sqrt{((-7)-0)^{2}-(-a)^{2}}} \right )= 0

V2 – V1 = 0

Work done = Charge (q) x Change in Potential (V2 – V1)

Since the change in potential is zero, the work done is also zero.

The change in potential is independent of the path taken between the two points. Therefore, the work done in moving a point charge will remain zero.

Q2. 22) Figure below 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 r/a >> 1, and contrast your results with that due to an electric dipole, and an electric monopole (i.e., a single charge).

NCERT Class 12 - Chapter 2 - Image - Question 2.12

Solution:

image name

Four charges are placed at points A,B,B and C respectively.

Let us consider a point P located at the axis of the quadrupole.

It can be considered that the electric quadrupole has three charges.

The charge +q is placed at A

The charge -2q is placed at B

The charge +q is placed at C

AB = BC = a

BP = r

PA = r + a

PZ = r – a

Therefore, the electrostatic potential due to the system of three charges is

V=14πϵ0[qPA2qPB+qPC]V =\frac{1}{4\pi \epsilon _{0}}\left [ \frac{q}{PA}-\frac{2q}{PB}+\frac{q}{PC} \right ] =14π0[qr+a2qr+qra]=q4π0[r(ra)2(r+a)(ra)+r(r+a)r(r+a)(ra)]=q4π0[r2ra2r2+2a2+r2+rar(r2a2)]=q4π0[2a2r(r2a2)]=2qa24π0r3(1a2r2)\begin{array}{l} =\frac{1}{4 \pi \in_{0}}\left[\frac{q}{r+a}-\frac{2 q}{r}+\frac{q}{r-a}\right] \\ =\frac{q}{4 \pi \in_{0}}\left[\frac{r(r-a)-2(r+a)(r-a)+r(r+a)}{r(r+a)(r-a)}\right] \\ =\frac{q}{4 \pi \in_{0}}\left[\frac{r^{2}-r a-2 r^{2}+2 a^{2}+r^{2}+r a}{r\left(r^{2}-a^{2}\right)}\right]=\frac{q}{4 \pi \in_{0}}\left[\frac{2 a^{2}}{r\left(r^{2}-a^{2}\right)}\right] \\ =\frac{2 q a^{2}}{4 \pi \in_{0} r^{3}\left(1-\frac{a^{2}}{r^{2}}\right)} \end{array}

Since r/a >>1,

a/r<< 1

Therefore, a2/r2 is negligible

So we get,

V=2qa24πϵ0r3V = \frac{2qa^{2}}{4\pi \epsilon _{0}r^{3}}

Therefore we get,

V∝ 1/r3

However, for a dipole, V ∝ 1/r2

And for a monopole, V ∝ 1/r

Q2.23) An electrical technician requires a capacitance of 2 µF in a circuit across a potential difference of 1 kV. A large number of 1 µF capacitor are available to him each of which can withstand a potential difference of not more than 400 V. Suggest a possible arrangement that requires the minimum number of capacitors.

Solution:

Required Capacitance, C= 2μF

Potential difference, V = 1 kV = 1000 V

Capacitance of each capacitor, C1 = 1μF

Potential difference that the capacitors can withstand, V1 = 400 V

Suppose a number of capacitors are connected in series and then connected parallel to each other. Then the number of capacitors in each row is given by

1000/400 = 2.5

Therefore, the number of capacitors connected in series is three.

So the capacitance of each row is 11+1+1=13μF\frac{1}{1+1+1}= \frac{1}{3}\mu F

Let there be n parallel rows. Each of these rows will have 3 capacitors. Therefore, the equivalent capacitance of the circuit is given as

13+13+13+——nterms\frac{1}{3}+\frac{1}{3}+\frac{1}{3}+——-n \, terms

The required capacitance of the circuit is 2μF

Therefore, n/3 = 2

n = 6

Therefore, there are 6 rows of three capacitors in the circuit. A minimum of 6 x 3 = 18 capacitors are required.

Q2.24) What is the area of the plates of a 2 F parallel plate capacitor, given that the separation between the plates is 0.5 cm? [You will realise from your answer why ordinary capacitors are in the range of µF or less. However, electrolytic capacitors do have a much larger capacitance (0.1 F) because of the very minute separation between the conductors.]

Solution:

Capacitance of the parallel plate capacitor, C = ε0A/d

Capacitance of the capacitor, C=2 F

Separation between the plates, d= 0.5 cm = 0.5 x 10-2 m

ε= permittivity of the free space = 8.85 x 10-12 C2N-1m-2

Area of the plates , A = Cd/ε0

A = [2 x 0.5 x 10-2]/8.85 x 10-12

= 1130 x 106 m2

Q2.25) Obtain the equivalent capacitance of the network in Figure. For a 300 V supply, determine the charge and voltage across each capacitor.

NCERT Class 12 - Chapter 2 - Image - Question 2.25

Solution:

The above figure can be redrawn as below

NCERT Class 12 - Chapter 2 - Image 2 - Question 2.25

The capacitors C2 and C3 are connected in series. The equivalent capacitance C’

1C=1C2+1C3=1200+1200=2200\frac{1}{C’}= \frac{1}{C_{2}}+\frac{1}{C_{3}} = \frac{1}{200}+\frac{1}{200}=\frac{2}{200}

C’ = 100 pF

The capacitors C’ and C1 are parallel. The equivalent capacitance is C” = C’ + C1

C” = 100 + 100 = 200 pF

C” and C4 are connected in series. Let the equivalent capacitance be C

1C=1C+1C4=1200+1100=3200\frac{1}{C} = \frac{1}{C^{”}}+\frac{1}{C_{4}} = \frac{1}{200}+\frac{1}{100}=\frac{3}{200}

C = 200/3 pF

Hence the equivalent capacitance of the circuit is 200/3 pF

Total charge, Q = CV = 2003×1012×300=2×108C\frac{200}{3}\times 10^{-12}\times 300 = 2 \times 10^{-8}C

Q = Q4 = 2 x 10-8 C

Potential difference across C4, V4 = Q/C4

= (2 x 10 -8)/(100 x 10-12) = 200 V

Potential difference across C”,  V” = 300 V – 200 V = 100 V

Potential difference across C1,  V1 = V” = 100 V

Charge across C1, Q1 = C1V1 = (100 x 10-12) x 100 = 10-8 C

The charge across C2 and C3, Q2 = Q – Q1 = 2 x 10-8 – 10-8 

= 10-8 C

Potential across C2 , V2 = Q2/C2 = 10-8/ (200 x 10-12) = 50 V

Potential across C3, V3 = Q2/C3 = 10-8/(200 x 10-12) = 50 V

Therefore,

Q1 = 10-8 C , V1 = 100 V

Q2 =  10-8 C, V2 = 50 V

Q2 = Q3 = 10-8 C, V3 = 50 V

Q4 = 2 x 10-8 C , V4 = 200 V

Q2.26) The plates of a parallel plate capacitor have an area of 90 cm2 each and are separated by 2.5 mm. The capacitor is charged by connecting it to a 400 V supply.
(a) How much electrostatic energy is stored by the capacitor?
(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

Solution:

Area of the plates of a parallel plate capacitor, A = 90 cm2 = 90 x 10-4 m2

Separation between the plates, d = 2.5 mm = 2.5 x 10-3 m

Potential difference across the plates, V = 400 V

Capacitance of the capacitor, C = ε0A/d

Electrostatic energy stored in the capacitor, E = (1/2) CV2

=12ϵ0AdV2= \frac{1}{2}\frac{\epsilon _{0}A}{d}V^{2} =128.85×1012×90×104×40022×2.5×103=2.55×106J= \frac{1}{2}\frac{8.85 \times 10^{-12}\times 90\times 10^{-4}\times 400^{2}}{2\times 2.5\times 10^{-3}} = 2.55 \times 10^{-6}J

Therefore, the electrostatic energy stored by the capacitor is 2.55 x 10-6 J

Volume of the capacitor, V = A x d

= 90 x 10-4  x 25 x 10-3

= 2.25 x 10-4 m3

Energy stored in the capacitor per unit volume is

u = E/V = 12CV2Ad=ϵ0A2dV2Ad=12ϵ0(Vd)2\frac{\frac{1}{2}CV^{2}}{Ad} = \frac{\frac{\epsilon _{0}A}{2d}V^{2}}{Ad}= \frac{1}{2}\epsilon _{0}(\frac{V}{d})^{2}

Here, V/d = Electric intensity = E

Therefore, u=12ϵ0E2u = \frac{1}{2}\epsilon _{0}E^{2}

Q2.27) A 4 µF capacitor is charged by a 200 V supply. It is then disconnected from the supply and is connected to another uncharged 2 µF capacitors. How much electrostatic energy of the first capacitor is lost in the form of heat and electromagnetic radiation?

Solution:

Capacitance of the capacitor, C1= 4 μF

Supply voltage, V1 = 200 V

Capacitance of the uncharged capacitor, C2= 2 μF

Electrostatic energy stored in C1 is given as

E1 = (1/2)C1V12

= (1/2) x 4 x 10-6 x (200)2

= 8 x 10-2 J

When C1 is disconnected from the power supply and connected to C2, the voltage acquired by it is V2.

According to the law of conservation of energy, the initial charge on the capacitor C1 is equal to the final charge on the capacitors C1 and C2.

V2 (C1 + C2) = C1V1

V2 (4+ 2) x 10-6 =  4 x 10-6 x 200

V2 = (400/3) V

Electrostatic energy of the combination is

E2 =  (1/2)(C1+C2) V12

= (1/2) x (2+4) x 10-6 x (400/3)2

= 5.33 x 10-2 J

Hence, amount of electrostatic energy lost by capacitor C1 = E1 – E2

= 0.08 – 0.0533 = 0.0267

= 2.67 x 10-2 J

Q2.28) Show that the force on each plate of a parallel plate capacitor has a magnitude equal to (½) 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 ½.

Solution:

Let F be the force required to separate the plates of the parallel plate capacitors.

Let Δx be the distance between the two plates.

Therefore, the work done to separate the plates, W = F Δx. As a result, the potential energy of the capacitor increases by an amount equal to uAΔx.

here, u = energy density

A = area of each plate

d = distance between the plates

V = potential difference across the plates

The work done will be equal to the increase in potential energy

FΔx = uAΔx

F = uA

Substituting u=12ϵ0E2u = \frac{1}{2}\epsilon _{0}E^{2} in above equation

F=uA=12ϵ0E2AF= uA = \frac{1}{2}\epsilon _{0}E^{2}A

The electric intensity, E = V/d

F=uA=12ϵ0VdEAF= uA = \frac{1}{2}\epsilon _{0}\frac{V}{d}EA

However, capacitance, C=ϵ0AdC = \frac{\epsilon _{0}A}{d}

Therefore, F = (1/2) (CV) E

Charge on the capacitor is given as Q = CV

Therefore, F = (1/2)QE

The electric field just outside the conductor is E and inside the conductor is zero. Hence, the average of the electric field E/2 is equal to the force.

Q2.29) A spherical capacitor consists of two concentric spherical conductors, held in position by suitable insulating supports (as shown in the figure). Show that the capacitance of a spherical capacitor is given by

C=4πϵ0r1r2r1r2C = \frac{4\pi \epsilon _{0}r_{1}r_{2}}{r_{1}-r_{2}}