Given below are two statements. Statement I : Electric potential is constant within and at the surface of each conductor. Statement II : Electric field just outside a charged conductor is perpendicular to the surface of the conductor at every point. In the light of the above statements, choose the most appropriate answer from the options given below.

Both statement I and statement II are correct

Statement I is correct but statement II is correct.

Statement I is incorrect but statement II is correct.

Class 12 Physics Chapter 2 Electrostatic Potential And Capacitance Solutions: Free PDF and Important Formulas

Q: 2.1 Two charges 5 × 10–8 C and –3 × 10–8 C are located 16 cm 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.

2.1 Let the charges be q1 = 5 ×10-8 C and q2 = -3 ×10-8 C Distance between the two charges, d = 16 cm = 0.16 m Consider a point P on the line joining the two charges R = distance of the point P from the charge q1 Let the electrical potential (V) at point P is zero Potential at point P caused by charges q1 and q2 respectively. V = 14πε0×q1r + 14πε0×q2(d-r) …………………………..(1) Where ε0 = permittivity of free space For V = 0, the equation (1) becomes 14πε0×q1r=-14πε0×q2(d-r) or q1r = -q2(d-r) or 5×10-8r = --3×10-8(d-r) 5×10-8r = 3×10-8(d-r) or r(d-r) = 5×10-83×10-8 or r(d-r) = 53 3r = 5d -5r or r = (5d/8) = 0.1 m = 10 cm Therefore the potential is zero at a distance of 10 cm from the charge q1 Suppose the point P is outside the system of two charges, as shown in the following figure. V = 14πε0×q1s + 14πε0×q2(s-d) ……………………………(2) For V = 0, we get 14πε0×q1s=-14πε0×q2(s-d) or q1s=-q2(s-d) or 5×10-8s=--3×10-8(s-d) 5s = 3s-0.16 or 5s – 0.8 = 3s or s = 0.82 = 0.4 m = 40 cm Therefore, the potential is zero at a distance 40 cm from q1 , outside the system of charge.

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.

2.2 The charge at each corner, q1 = 5 μC=5×10-6 C The sides of the hexagon, AB = BC = CD = DE = EF =FA = 10 cm = 0.1 m Let O be the centre. The electric potential at O is given by V = 14πε0×6q1d , where ε0 = permittivity of free space = 8.854 ×10-12 C2N-1 m-2 Hence V = 14×π×8.854×10-12×6×5×10-60.1 = 2.696 ×106 V Therefore, the potential at the centre is 2.7 ×106 V

Q: 2.3 Two charges 2 μC and –2 μC are placed at points A and B, 6 cm apart. (a) Identify an equipotential surface of the system. (b) What is the direction of the electric field at every point on this surface?

2.3 The arrangement is represented in the adjoining figure. An equipotential surface is the plane on which total potential is zero everywhere. The plane is normal to line AB. The plane is located at the midpoint of the line AB because the magnitude of the charge is same. The direction of the electric field at every point on that surface 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 × 10–7C distributed uniformly on its surface. What is the electric field? (a) Inside the sphere (b) Just outside the sphere (c) At a point 18 cm from the centre of the sphere?

2.4 Radius of the spherical conductor, r = 12 cm = 0.12 m Uniformly distributed charge, q = 1.6 ×10-7 C The electrical field inside the conductor is zero. Electric field E just outside the conductor is given by E = 14πε0×qr2 where ε0 = permittivity of free space = 8.854 ×10-12 C2N-1 m-2 E = 14×π×8.854×10-12×1.6×10-70.122 = 99863.8 N C-1 ≅105 N C-1 At a point 18 cm from the centre of the sphere. Hence r = 18 cm = 0.18 m From the above equation we get E at 18 cm from the centre = 14×π×8.854×10-12×1.6×10-70.182 = 4.438 ×104 N C-1

Q: 2.5 A parallel plate capacitor with air between the plates has a capacitance of 8 pF (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?

2.5 Capacitance between the parallel plates of the capacitor, C = 8 pF Let us assume, initially, 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 = kε0Ad = ε0Ad , since k = 1……………………..(1) Where A = area of each plate ε0 = permittivity of free space = 8.854 ×10-12 C2N-1 m-2 d = distance between two plates If the distance between two plates is reduced to half, then distance between two plates d1 = d2 Dielectric constant of a new substance, k1 = 6 Then the resistance between two plates, C1 = 6ε0Ad/2 = 12ε0Ad ……………(2) From equation (1) and (2) we get C1=12C = 12 ×8 pF = 96 pF

Updated on Oct 14, 2025 13:41 IST

By nitesh singh, Senior Executive

This chapter is another major milestone in Electrostatics. Chapter 2 of Class 12 Physics mainly deals with the concept of energy stored due to static charges. This chapter holds significant weightage in boards and other competitive exams. Find NCERT solutions for Electrostatic Potential and Capacitance with complete questions and answers.

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Electrostatic Potential and Capacitance Solutions

Q. 2.1 Two charges 5 × 10^–8 C and –3 × 10^–8 C are located 16 cm 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.2.1

Let the charges be

$q_{1}$ = 5 $\times 10^{- 8}$ C and $q_{2}$ = -3 $\times 10^{- 8}$ C

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

Consider a point P on the line joining the two charges

R = distance of the point P from the charge $q_{1}$

Let the electrical potential (V) at point P is zero

Potential at point P caused by charges $q_{1}$ and $q_{2}$ respectively.

V = $\frac{1}{4 π ε_{0}} \times \frac{q_{1}}{r}$ $+$ $\frac{1}{4 π ε_{0}} \times \frac{q_{2}}{(d - r)}$ …………………………..(1)

Where $ε_{0}$ = permittivity of free space

For V = 0, the equation (1) becomes

$\frac{1}{4 π ε_{0}} \times \frac{q_{1}}{r} = - \frac{1}{4 π ε_{0}} \times \frac{q_{2}}{(d - r)}$ or $\frac{q_{1}}{r}$ = $- \frac{q_{2}}{(d - r)}$ or $\frac{5 \times 10^{- 8}}{r}$ = $- \frac{- 3 \times 10^{- 8}}{(d - r)}$

$\frac{5 \times 10^{- 8}}{r}$ = $\frac{3 \times 10^{- 8}}{(d - r)}$ or $\frac{r}{(d - r)}$ = $\frac{5 \times 10^{- 8}}{3 \times 10^{- 8}}$ or $\frac{r}{(d - r)}$ = $\frac{5}{3}$

3r = 5d -5r or r = (5d/8) = 0.1 m = 10 cm

Therefore the potential is zero at a distance of 10 cm from the charge $q_{1}$

Suppose the point P is outside the system of two charges, as shown in the following figure.

V = $\frac{1}{4 π ε_{0}} \times \frac{q_{1}}{s}$ $+$ $\frac{1}{4 π ε_{0}} \times \frac{q_{2}}{(s - d)}$ ……………………………(2)

For V = 0, we get

$\frac{1}{4 π ε_{0}} \times \frac{q_{1}}{s} = - \frac{1}{4 π ε_{0}} \times \frac{q_{2}}{(s - d)}$ or $\frac{q_{1}}{s} = - \frac{q_{2}}{(s - d)}$ or $\frac{5 \times 10^{- 8}}{s} = - \frac{- 3 \times 10^{- 8}}{(s - d)}$

$\frac{5}{s}$ = $\frac{3}{s - 0.16}$ or 5s – 0.8 = 3s or s = $\frac{0.8}{2}$ = 0.4 m = 40 cm

Therefore, the potential is zero at a distance 40 cm from $q_{1}$ , outside the system of charge.

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.

Ans.2.2

The charge at each corner, $q_{1}$ = 5 $μ C = 5 \times 10^{- 6}$ C

The sides of the hexagon, AB = BC = CD = DE = EF =FA = 10 cm = 0.1 m

Let O be the centre. The electric potential at O is given by

V = $\frac{1}{4 π ε_{0}} \times \frac{6 q_{1}}{d}$ , where $ε_{0}$ = permittivity of free space = 8.854 $\times 10^{- 12}$ $C^{2} N^{- 1}$ $m^{- 2}$

Hence V = $\frac{1}{4 \times π \times 8.854 \times 10^{- 12}} \times \frac{6 \times 5 \times 10^{- 6}}{0.1}$ = 2.696 $\times 10^{6}$ V

Therefore, the potential at the centre is 2.7 $\times 10^{6}$ V

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

(a) Identify an equipotential surface of the system.

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

Ans.2.3 The arrangement is represented in the adjoining figure.

An equipotential surface is the plane on which total potential is zero everywhere. The plane is normal to line AB. The plane is located at the midpoint of the line AB because the magnitude of the charge is same.

The direction of the electric field at every point on that surface 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 × 10^–7C distributed uniformly on its surface. What is the electric field?

(a) inside the sphere

(b) just outside the sphere

Ans.2.4 Radius of the spherical conductor, r = 12 cm = 0.12 m

Uniformly distributed charge, q = 1.6 $\times 10^{- 7}$ C

The electrical field inside the conductor is zero.

Electric field E just outside the conductor is given by

E = $\frac{1}{4 π ε_{0}} \times \frac{q}{r^{2}}$ where $ε_{0}$ = permittivity of free space = 8.854 $\times 10^{- 12}$ $C^{2} N^{- 1}$ $m^{- 2}$

E = $\frac{1}{4 \times π \times 8.854 \times 10^{- 12}} \times \frac{1.6 \times 10^{- 7}}{{0.12}^{2}}$ = 99863.8 N $C^{- 1}$ $≅ 10^{5}$ N $C^{- 1}$

At a point 18 cm from the centre of the sphere. Hence r = 18 cm = 0.18 m

From the above equation we get

E at 18 cm from the centre = $\frac{1}{4 \times π \times 8.854 \times 10^{- 12}} \times \frac{1.6 \times 10^{- 7}}{{0.18}^{2}}$ = 4.438 $\times 10^{4}$ N $C^{- 1}$

Commonly asked questions

2.1 Two charges 5 × 10^–8 C and –3 × 10^–8 C are located 16 cm 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.

2.1

Let the charges be

$q_{1}$ = 5 $\times 10^{- 8}$ C and $q_{2}$ = -3 $\times 10^{- 8}$ C

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

Consider a point P on the line joining the two charges

R = distance of the point P from the charge $q_{1}$

Let the electrical potential (V) at point P is zero

Potential at point P caused by charges $q_{1}$ and $q_{2}$ respectively.

V = $\frac{1}{4 π ε_{0}} \times \frac{q_{1}}{r}$ $+$ $\frac{1}{4 π ε_{0}} \times \frac{q_{2}}{(d - r)}$ …………………………..(1)

Where $ε_{0}$ = permittivity of free space

For V = 0, the equation (1) becomes

$\frac{5 \times 10^{- 8}}{r}$ = $\frac{3 \times 10^{- 8}}{(d - r)}$ or $\frac{r}{(d - r)}$ = $\frac{5 \times 10^{- 8}}{3 \times 10^{- 8}}$ or $\frac{r}{(d - r)}$ = $\frac{5}{3}$

3r = 5d -5r or r = (5d/8) = 0.1 m = 10 cm

Therefore the potential is zero at a distance of 10 cm from the charge $q_{1}$

Suppose the point P is outside the system of two charges, as shown in the following figure.

V = $\frac{1}{4 π ε_{0}} \times \frac{q_{1}}{s}$ $+$ $\frac{1}{4 π ε_{0}} \times \frac{q_{2}}{(s - d)}$ ……………………………(2)

For V = 0, we get

$\frac{5}{s}$ = $\frac{3}{s - 0.16}$ or 5s – 0.8 = 3s or s = $\frac{0.8}{2}$ = 0.4 m = 40 cm

Therefore, the potential is zero at a distance 40 cm from $q_{1}$ , outside the system of charge.

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.

2.2

The charge at each corner, $q_{1}$ = 5 $μ C = 5 \times 10^{- 6}$ C

The sides of the hexagon, AB = BC = CD = DE = EF =FA = 10 cm = 0.1 m

Let O be the centre. The electric potential at O is given by

V = $\frac{1}{4 π ε_{0}} \times \frac{6 q_{1}}{d}$ , where $ε_{0}$ = permittivity of free space = 8.854 $\times 10^{- 12}$ $C^{2} N^{- 1}$ $m^{- 2}$

Hence V = $\frac{1}{4 \times π \times 8.854 \times 10^{- 12}} \times \frac{6 \times 5 \times 10^{- 6}}{0.1}$ = 2.696 $\times 10^{6}$ V

Therefore, the potential at the centre is 2.7 $\times 10^{6}$ V

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

(a) Identify an equipotential surface of the system.

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

2.3 The arrangement is represented in the adjoining figure.

The direction of the electric field at every point on that surface is normal to the plane in the direction of AB.

2.4 A spherical conductor of radius 12 cm has a charge of 1.6 × 10^–7C distributed uniformly on its surface. What is the electric field?

(a) Inside the sphere

(b) Just outside the sphere

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

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

Uniformly distributed charge, q = 1.6 $\times 10^{- 7}$ C

The electrical field inside the conductor is zero.

Electric field E just outside the conductor is given by

E = $\frac{1}{4 π ε_{0}} \times \frac{q}{r^{2}}$ where $ε_{0}$ = permittivity of free space = 8.854 $\times 10^{- 12}$ $C^{2} N^{- 1}$ $m^{- 2}$

E = $\frac{1}{4 \times π \times 8.854 \times 10^{- 12}} \times \frac{1.6 \times 10^{- 7}}{{0.12}^{2}}$ = 99863.8 N $C^{- 1}$ $≅ 10^{5}$ N $C^{- 1}$

At a point 18 cm from the centre of the sphere. Hence r = 18 cm = 0.18 m

From the above equation we get

E at 18 cm from the centre = $\frac{1}{4 \times π \times 8.854 \times 10^{- 12}} \times \frac{1.6 \times 10^{- 7}}{{0.18}^{2}}$ = 4.438 $\times 10^{4}$ N $C^{- 1}$

2.5 A parallel plate capacitor with air between the plates has a capacitance of 8 pF (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?

2.5 Capacitance between the parallel plates of the capacitor, C = 8 pF

Let us assume, initially, 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 = $\frac{k ε_{0} A}{d}$ = $\frac{ε_{0} A}{d}$ , since k = 1……………………..(1)

Where A = area of each plate

$ε_{0}$ = permittivity of free space = 8.854 $\times 10^{- 12}$ $C^{2} N^{- 1}$ $m^{- 2}$

d = distance between two plates

If the distance between two plates is reduced to half, then distance between two plates $d_{1}$ = $\frac{d}{2}$

Dielectric constant of a new substance, $k_{1}$ = 6

Then the resistance between two plates, $C_{1}$ = $\frac{6 ε_{0} A}{d / 2}$ = $\frac{12 ε_{0} A}{d}$ ……………(2)

From equation (1) and (2) we get

$C_{1} = 12 C$ = 12 $\times 8$ pF = 96 pF

2.6 Three capacitors each of capacitance 9 pF are connected in series.

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

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

2.6 Capacitance of each of the three capacitors, C = 9 pF

The equivalent capacitance $C_{e q}$ when the capacitors are connected in series is given by

$\frac{1}{C_{e q}}$ = $\frac{1}{C}$ + $\frac{1}{C} + \frac{1}{C}$ = $\frac{3}{C}$ = $\frac{3}{9}$ = $\frac{1}{3}$

Hence, $C_{e q}$ = 3 pF

Supply voltage, V = 120 V

Potential difference ( $V_{1}$ ) across each capacitor is given by $V_{1}$ = $\frac{V}{3}$ = $\frac{120}{3}$ = 40 V

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

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

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

2.7 Let the three capacitors be

$C_{1}$ = 2 pF, $C_{2}$ = 3 pF and $C_{3}$ = 4 pF

The equivalent capacitance, $C_{e q}$ is given by

$C_{e q} = C_{1} + C_{2} + C_{3}$ = 2 + 3 + 4 = 9 pF

$(b)$ When the combination is connected to a 100 V supply, the voltage in all three capacitors = 100 V

$T h e$ charge in each capacitor is given by the relation, q = VC

Hence for $C_{1}$ , $q_{1}$ = 100 $\times 2 p C$ = 200 pC = 2 $\times 10^{- 10}$ C,

for $C_{2}$ , $q_{2}$ = 100 $\times 3 p C$ = 300 pC = 3 $\times 10^{- 10}$ C,

for $C_{3}$ , $q_{3}$ = 100 $\times 4 p C$ = 400 pC = 4 $\times 10^{- 10}$ C

2.8 In a parallel plate capacitor with air between the plates, each plate has an area of 6 × 10^–3 m² 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?

2.8 Area of each plate, A = 6 $\times 10^{- 3}$ $m^{2}$

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

Supply voltage, V = 100 V

Capacitance C of the parallel plate is given by C = $\frac{k ε_{0} A}{d}$

In case of air, dielectric constant k = 1 and

$ε_{0}$ = permittivity of free space = 8.854 $\times 10^{- 12}$ $C^{2} N^{- 1}$ $m^{- 2}$

Hence C = $\frac{8.854 \times 10^{- 12} \times 6 \times 10^{- 3}}{3 \times 10^{- 3}}$ F = 17.708 $\times 10^{- 12}$ F = 17.71 pF

Charge on each plate of the capacitor is given by q = VC = 100 $\times 17.71 \times 10^{- 12}$ C

= 1.771 $\times 10^{- 9}$ C

Therefore, capacitance of the capacitor is 17.71 pF and charge on each plate is 1.77 $\times 10^{- 9}$ C

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.

2.9 Dielectric constant of the mica sheet, k =6

While the voltage supply remained connected :

V = 100 V

Initial capacitance, C = 17.708 $* 10^{- 12}$ F

New capacitance, $C_{1}$ = kC = 6 $*$ 17.708 $* 10^{- 12} F$ = 106.25 $* 10^{- 12}$ F

= 106.25 pF

New charge, $q_{1}$ = $C_{1} V$ = 106.25 $* 10^{- 12} * 100$ C = 10.62 $* 10^{- 9}$ C

If the supply voltage is removed, then there will be constant amount of charge in the plates.

Charge, q = CV = 17.708 $* 10^{- 12} * 100$ C = 1.7708 $* 10^{- 9}$ C

Potential across plates, $V_{1}$ = $\frac{q}{C_{1}}$ = $\frac{1.7708 * 10^{- 9}}{106.25 * 10^{- 12}}$ = 16.66 V

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?

2.11 Capacitance of the first capacitor, C = 600 pF = 600 $\times 10^{- 12}$ F

Potential difference, V = 200V

Electrostatic energy of the first capacitor, $E_{1}$ = $\frac{1}{2}$ C $V^{2}$ = $\frac{1}{2} \times$ 600 $\times 10^{- 12} \times 200^{2}$ J

= 1.2 $\times 10^{- 5}$ J

After the supply is disconnected and the first capacitor is connected to another capacitor of 600 pF capacitance in series, the equivalent capacitance $C_{e q}$ is given by

$\frac{1}{C_{e q}}$ = $\frac{1}{c}$ + $\frac{1}{c}$ = $\frac{1}{600}$ + $\frac{1}{600}$ = $\frac{1}{300}$

$C_{e q} = 300 p F$

New electrostatic energy, $E_{2}$ = $\frac{1}{2} C_{e q} V^{2}$ = $\frac{1}{2} \times 300 \times 10^{- 12} \times {(200)}^{2}$ = 6 $\times 10^{- 6}$ J

Electrostatic energy lost = $E_{1} - E_{2}$ = 1.2 $\times 10^{- 5}$ $-$ 6 $\times 10^{- 6}$ = 6 $\times 10^{- 6}$ J

2.12 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).

2.12

Charge located at the origin O, q=8 mC= 8 $\times 10^{- 3}$ C

Magnitude of a small charge, which is taken from point P to Q, $q_{1}$ = -2 $\times 10^{- 9}$ C

Pont P is at a distance, $d_{1}$ = 3 cm = 0.03 m from origin, along Z axis

Point Q is at a distance, $d_{2}$ = 4 cm = 0.04 m from origin, along y axis

Potential at point P, $V_{1}$ = $\frac{q}{4 π ε_{0} d_{1}}$

Potential at point Q, $V_{2}$ = $\frac{q}{4 π ε_{0} d_{2}}$

$ε_{0}$ = permittivity of free space = 8.854 $\times 10^{- 12}$ $C^{2} N^{- 1}$ $m^{- 2}$

Work done, W = $q_{1}$ $(V_{2} - V_{1})$ = $\frac{q q_{1}}{4 π ε_{0}} (\frac{1}{d_{2}} - \frac{1}{d_{1}})$ = $\frac{8 \times 10^{- 3} \times - 2 \times 10^{- 9}}{4 \times π \times 8.854 \times 10^{- 12}} (\frac{1}{0.04} - \frac{1}{0.03})$

= (-0.1438) $\times (- 8.333)$ = 1.198 J

2.13 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.

2.13

Length of the side of a cube = b

Charge at each vertices= q

d = diagonal of each face, d = $\sqrt{b^{2} + b^{2}}$ = b $\sqrt 2$

l = length of the solid diagonal, l = $\sqrt{d^{2} + b^{2}}$ = b $\sqrt 3$

If r is the distance from the centre of the cube to its corner, then r = l/2 = $\frac{\sqrt 3}{2} b$

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

V = $\frac{8 q}{4 π ε_{0} r}$ = $\frac{8 q}{4 π ε_{0} \frac{\sqrt 3}{2} b}$ = $\frac{4 q}{\sqrt 3 π ε_{0} b}$

Therefore, the potential at the centre of the cube is $\frac{4 q}{\sqrt 3 π ε_{0} b}$

The electric field at the centre of the cube, due to eight charges, gets cancelled, since the charges are distributed symmetrically. The electric field is zero at the centre.

2.14 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.

2.14

Two charges are at point A and B. O is the midpoint of line joining A & B

Magnitude of charge at point A, $q_{1}$ = 1.5 $μ C$ = 1.5 $\times 10^{- 6} C$ and

at point B, $q_{2}$ = 2.5 $μ C = 2.5 \times 10^{- 6} C$

Distance between A & B, d = 30 cm = 0.3 m

Let $V_{1}$ and $E_{1}$ be the electric potential and electric field respectively at O.

$V_{1} =$ Potential due to charge at A + Potential due to charge at B

$= \frac{q_{1}}{4 π ε_{0} \frac{d}{2}}$ + $\frac{q_{2}}{4 π ε_{0} \frac{d}{2}}$ = $\frac{1}{4 π ε_{0} \frac{d}{2}}$ ( $q_{1} + q_{2})$

$ε_{0}$ = permittivity of free space = 8.854 $\times 10^{- 12}$ $C^{2} N^{- 1}$ $m^{- 2}$

$V_{1} =$ $\frac{1}{4 π \times 8.854 \times 10^{- 12} \times \frac{0.3}{2}}$ (1.5 $\times 10^{- 6} + 2.5 \times 10^{- 6})$ = 2.4 $\times 10^{5}$ V

$E_{1} =$ Electric field due to $q_{2}$ - Electric field due to $q_{1}$

$=$ $\frac{q_{2}}{4 π ε_{0} ({\frac{d}{2})}^{2}}$ $-$ $\frac{q_{1}}{4 π ε_{0} ({\frac{d}{2})}^{2}}$

$=$ $\frac{1}{4 π ε_{0} ({\frac{d}{2})}^{2}} (q_{2} - q_{1})$ = $\frac{1}{4 π \times 8.854 \times 10^{- 12} \times ({\frac{0.3}{2})}^{2}} (2.5 \times 10^{- 6} - 1.5 \times 10^{- 6})$

= 4.0 $\times 10^{5}$ V/m

Therefore, the potential at mid-point is 2.4 $\times 10^{5}$ V and the electric field is

4.0 $\times 10^{5}$ V/m and it is directed from larger charge to smaller charge.

Let us consider a point Z such that normal distance OZ = 10 cm = 0.1 m

Let $V_{2}$ and $E_{2}$ be the electric potential and electric field respectively at Z.

From the $?$ ABZ, we can write,

BZ = AZ = $\sqrt{{0.15}^{2} + {0.1}^{2}}$ = 0.18 m

$V_{2} =$ Potential due to charge at A + Potential due to charge at B

$= \frac{q_{1}}{4 π ε_{0} A Z}$ + $\frac{q_{2}}{4 π ε_{0} B Z}$ = $\frac{1}{4 π ε_{0} 0.18}$ ( $q_{1} + q_{2})$

$ε_{0}$ = permittivity of free space = 8.854 $\times 10^{- 12}$ $C^{2} N^{- 1}$ $m^{- 2}$

$V_{2} =$ $\frac{1}{4 π \times 8.854 \times 10^{- 12} \times 0.18}$ (1.5 $\times 10^{- 6} + 2.5 \times 10^{- 6})$ = 2.0 $\times 10^{5}$ V

Electric field due to charge $q_{1} a t Z, E_{A}$

$E_{A} = \frac{q_{1}}{4 π ε_{0} ({A Z)}^{2}}$ = $\frac{1.5 \times 10^{- 6}}{4 \times π \times 8.854 \times 10^{- 12} \times ({0.18)}^{2}}$ = 0.416 $\times 10^{6}$ V/m

Electric field due to charge $q_{2} a t Z, E_{B}$

$E_{B} = \frac{q_{2}}{4 π ε_{0} ({B Z)}^{2}}$ = $\frac{2.5 \times 10^{- 6}}{4 \times π \times 8.854 \times 10^{- 12} \times ({0.18)}^{2}}$ = 0.693 $\times 10^{6}$ V/m

The resultant intensity at Z

$E_{2}$ = $\sqrt{{E_{A}}^{2} + {E_{B}}^{2} + 2 E_{A} E_{B} c o s 2 θ}$ where 2 $θ$ is the $∠ A Z B$

From the figure, we get $\cos ? θ$ = $\frac{Z O}{Z B}$ = $\frac{0.1}{0.18}$

Hence $θ$ = 56.25 $°$

Therefore, $E_{2}$ = $E_{2} = \sqrt{({0.416 \times 10^{6})}^{2} + {(0.693 \times 10^{6})}^{2} + 2 {(0.416 \times 10^{6}) \times (0.693 \times 10^{6})}_{} c o s 2 \times 56.25 °}$ = 6.57 $\times 10^{5}$ V/m

2.15 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.

2.15 Charge placed at the centre of the shell is +q. Hence a charge of magnitude of –q will be induced to the inner surface of the shell. Therefore, the total charge on the inner surface of the shell is –q.

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

$σ_{1} = \frac{T o t a l c h a r g e}{I n n e r s u r f a c e a r e a} = \frac{- q}{4 π r_{1}^{2}}$ …………(1)

When a charge of magnitude of Q is placed on the outer shell and the radius of outer shell is $r_{2}$ . The total charge experienced by the outer shell will be Q + q. The surface charge density at the outer surface of the shell is given by

$σ_{2} = \frac{T o t a l c h a r g e}{O u t e r s u r f a c e a r e a} = \frac{Q + q}{4 π r_{2}^{2}}$ …………(2)

The electric field intensity inside a cavity is zero, even if the shell is not spherical and has any irregular shape.

(a) Show that the normal component of electrostatic field has a discontinuity from one side of a charged surface to another given by

$(E_{2}$ - $E_{1}) . \hat{n} = \frac{σ}{ε_{0}}$

Where $\hat{n}$ is a unit vector normal to the surface at a point and s 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 $σ \hat{n}$ / $ε_{0}$ .

(b) Show that the tangential component of 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.]

2.16 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 infinite plane charged body has a uniform thickness, then electric field due to one surface of the charged body is given by,

$\overset{?}{E_{1}}$ = $\frac{σ}{2 ε_{0}} \overset{?}{n}$ ………….(i)

Where,

$\overset{?}{n}$ = unit vector normal to the surface at a point

$σ$ = surface charge density at that point

Electric field due to the other surface of the charged body is given by

$\overset{?}{E_{2}}$ = $\frac{σ}{2 ε_{0}} \overset{?}{n}$ ………….(ii)

$E l e c t r i c f i e l d a t a n y p o i n t$ due to the two surfaces,

$\overset{?}{E_{2}} - \overset{?}{E_{1}} = \frac{σ}{2 ε_{0}} \overset{?}{n} + \frac{σ}{2 ε_{0}} \overset{?}{n} = \frac{σ}{ε_{0}} \overset{?}{n}$ ……(iii)

$S i n c e i n s i d e a c l o s e d c o n d u c t o r, \overset{?}{E_{1}} = 0,$

$\overset{?}{E}$ = $\overset{?}{E_{2}}$ = $\frac{σ}{ε_{0}} \overset{?}{n}$

Therefore, the electric field just outside the conductor is $\frac{σ}{ε_{0}} \overset{?}{n}$

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 electrostatic field is continuous from one side of a charged surface to the other.

2.17 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?

2.17

Let charge density of the long charged cylinder, with length L and radius r be $λ .$

Another cylinder of same length surrounds the previous cylinder with radius R.

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

Electric flux through the Gaussian surface is given by Gaussian’s theorem as

$φ = E (2 π$ dL) = $\frac{q}{ε_{0}}$ , where

$q = c h a r g e o n t h e i n n e r s p h e r e o f t h e o u t e r c y l i n d e r$

$ε_{0} = P e r m i t t i v i t y o f f r e e s p a c e$

Then, $φ = E (2 π$ dL) = $\frac{q}{ε_{0}}$ = $\frac{λ L}{ε_{0}}$

E = $\frac{λ}{{2 π d ε}_{0}}$

Therefore, the electric field in the space between the two cylinders is $\frac{λ}{{2 π d ε}_{0}} .$

2.18 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 separations?

2.18 The distance between electron – proton of a hydrogen atom, d = 0.53 Å = 0.53 $\times 10^{- 10}$ m

Charge of an electron, $q_{1}$ = $-$ 1.6 $\times 10^{- 19}$ C

Charge of a proton, $q_{2}$ = 1.6 $\times 10^{- 19}$ C

Potential at infinity = 0

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

= 0 - $\frac{q_{1} q_{2}}{4 π ε_{0} d}$ , where $ε_{0} = P e r m i t t i v i t y o f f r e e s p a c e$ == 8.854 $\times 10^{- 12}$ $C^{2} N^{- 1}$ $m^{- 2}$

= 0 - $\frac{1.6 \times 10^{- 19} \times 1.6 \times 10^{- 19}}{4 π \times 8.854 \times 10^{- 12} \times 0.53 \times 10^{- 10}}$ = - 4.34 $\times 10^{- 18}$ J = $\frac{4.34 \times 10^{- 18}}{1.6 \times 10^{- 19}}$ = - 27.13 eV

(Since 1 eV = 1.6 $\times 10^{- 19}$ J)

Kinetic energy = $\frac{1}{2}$ of potential energy = $\frac{1}{2}$ $\times -$ 27.13 eV = -13.57 eV

Total energy = - 13.57 – (-27.13) = 13.57 eV

Therefore, the minimum work required to free the electron is 13.57 eV

In case of zero potential energy, $d_{1}$ = 1.06 Å = 1.06 $\times 10^{- 10}$ m

Potential energy of the system = Potential energy at $d_{1}$ - Potential energy at d

= $\frac{q_{1} q_{2}}{4 π ε_{0} d}$ $\times \frac{1}{1.6 \times 10^{- 19}}$ - 27.13 eV = $\frac{{(1.6 \times 10^{- 19})}^{2}}{4 π \times 8.854 \times 10^{- 12} \times 1.06 \times 10^{- 10}} \times \frac{1}{1.6 \times 10^{- 19}} -$ 27.13 = 13.56 -27.13 eV= -13.56 eV

2.19 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 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.

2.19

Let the charge in Proton 1 be $q_{1}$ = 1.6 $\times 10^{- 19}$ C, in Proton 2, $q_{2}$ = 1.6 $\times 10^{- 19}$ C and the charge in the electron, $q_{3}$ = $-$ 1.6 $\times 10^{- 19}$ C

Distance between Proton 1 & Proton 2, $d_{1}$ = 1.5 Å = 1.5 $\times 10^{- 10}$ m

Distance between Proton 1 and Electron, $d_{2}$ = 1.0 Å = 1.0 $\times 10^{- 10}$ m

Distance between Proton 2 and Electron, $d_{3}$ = 1.0 Å = 1.0 $\times 10^{- 10}$ m

The potential energy of the system,

V = $\frac{q_{1} q_{2}}{4 π ε_{0} d_{1}}$ + $\frac{q_{2} q_{3}}{4 π ε_{0} d_{3}}$ + $\frac{q_{3} q_{1}}{4 π ε_{0} d_{2}}$ ,

Where $ε_{0} = P e r m i t t i v i t y o f f r e e s p a c e$ = 8.854 $\times 10^{- 12}$ $C^{2} N^{- 1}$ $m^{- 2}$

V = $\frac{1}{4 π ε_{0}}$ ( $\frac{q_{1} q_{2}}{d_{1}}$ + $\frac{q_{2} q_{3}}{d_{2}}$ + $\frac{q_{3} q_{1}}{d_{2}}$ )

= $\frac{1}{4 π \times 8.854 \times 10^{- 12}} \times$ ( $\frac{1.6 \times 10^{- 19} \times 1.6 \times 10^{- 19}}{1.5 \times 10^{- 10}}$ + $\frac{1.6 \times 10^{- 19} \times - 1.6 \times 10^{- 19}}{1.0 \times 10^{- 10}}$ + $\frac{- 1.6 \times 10^{- 19} \times 1.6 \times 10^{- 19}}{1.0 \times 10^{- 10}}$ )

= 8.988 $\times 10^{9} \times (1.707 \times 10^{- 28}$ $- 2.56 \times 10^{- 28} - 2.56 \times 10^{- 28})$

= $- 3.067 \times 10^{- 18}$ J

= $- 3.067 \times 10^{- 18} \times \frac{1}{1.6 \times 10^{- 19}}$ eV

= - 19.17 eV

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

2.20 For Sphere A: We assume, radius = a, Charge on the sphere = $Q_{A}$ , Capacitance = $C_{A}$ , Electric field = $E_{A}$

For Sphere B: We assume, radius = b, Charge on the sphere = $Q_{B}$ , Capacitance = $C_{B}$ , Electric field = $E_{B}$

$E_{A} i s g i v e n b y, E_{A} =$ $\frac{Q_{A}}{4 π ε_{0} a^{2}}$ and $E_{B} i s g i v e n b y, E_{B} =$ $\frac{Q_{B}}{4 π ε_{0} b^{2}}$

$\frac{E_{A}}{E_{B}}$ = $\frac{Q_{A}}{Q_{B}} \times \frac{b^{2}}{a^{2}}$ ………..(1)

We also know Q = CV, hence $Q_{A}$ = $C_{A}$ V and $Q_{B}$ = $C_{B}$ V and $\frac{C_{A}}{C_{B}}$ = $\frac{a}{b}$

Then, $\frac{Q_{A}}{Q_{B}}$ = $\frac{C_{A}}{C_{B}} = \frac{a}{b}$ ……….(2)

Combining equations (1) and (2), we get

$\frac{E_{A}}{E_{B}}$ = $\frac{a}{b} \times \frac{b^{2}}{a^{2}}$ = $\frac{b}{a}$

2.21 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?

2.21 Charge –q is located at (0,0,-a) and charge +q is located at (0,0,a). Hence, they form a dipole. Point (0,0,z) is on the axis of this dipole and point (x,y,0) is normal to the axis of the dipole. Hence electrostatic potential at point (x,y,0) is zero. Electrostatic potential at point (0,0,z) is given by,

V = $\frac{1}{4 π ε_{0}}$ ( $\frac{q}{z - a}) + \frac{1}{4 π ε_{0}}$ ( $- \frac{q}{z + a})$ = $\frac{q (z + a - z + a)}{4 π ε_{0} (z^{2} - a^{2})}$ = $\frac{2 q a}{4 π ε_{0} (z^{2} - a^{2})}$ = $\frac{p}{4 π ε_{0} (z^{2} - a^{2})}$

Where $ε_{0}$ = Permittivity of free space

p = Dipole moment of the system of two charges = 2qa

Distance r is much greater than half of the distance between the two charges. Hence, the potential (V) at a distance r is inversely proportional to the distance. i.e. V $\propto$ $\frac{1}{r^{2}}$

Electrostatic potential ( $V_{1})$ at point (5,0,0) is given by

$V_{1}$ = $\frac{- q}{4 π ε_{0}} \frac{1}{\sqrt{{(5 - 0)}^{2} + {(- a)}^{2}}} + \frac{q}{4 π ε_{0}} \frac{1}{\sqrt{{(5 - 0)}^{2} + {(a)}^{2}}}$

= 0

Electrostatic potential ( $V_{2})$ at point (-7,0,0) is given by

$V_{2}$ = $\frac{- q}{4 π ε_{0}} \frac{1}{\sqrt{{(- 7)}^{2} + {(- a)}^{2}}} + \frac{q}{4 π ε_{0}} \frac{1}{\sqrt{{(- 7)}^{2} + {(a)}^{2}}}$

= 0

Hence, no work is done in moving a small test charge from point (5,0,0) to point (-7,0,0) along the x-axis.

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

2.22 Figure 2.32 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).

2.22 Four charges of same magnitudes are placed at points X, Y, Y and Z respectively, as shown in the figure.

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

Charge +q placed at point X

Charge -2q placed at Y

Charge +q placed at point Z.

XY = YZ = a

YP = r

PX = r + a

PZ = r – a

Electrostatic potential caused by the system of three charges at point P is given by,

V = $\frac{1}{4 π ε_{0}} [\frac{q}{X P}$ - $\frac{2 q}{Y P}$ + $\frac{q}{Z P}$ ]

= $\frac{1}{4 π ε_{0}} [\frac{q}{r + a}$ - $\frac{2 q}{r}$ + $\frac{q}{r - a}$ ]

= $\frac{q}{4 π ε_{0}} [\frac{r (r - a) - 2 (r - a) (r + a) + r (r + a)}{r (r - a) (r + a)}$ ]

= $\frac{q}{4 π ε_{0}} [\frac{r^{2} - r a - 2 r^{2} + 2 a^{2} + r^{2} + r a}{r (r - a) (r + a)}$ ]

= $\frac{q}{4 π ε_{0}} [\frac{2 a^{2}}{r (r^{2} - a^{2})}$ ]

= $\frac{2 q a^{2}}{4 π ε_{0} r^{3} (1 - \frac{a^{2}}{r^{2}})}$

Since $\frac{r}{a} ? 1,$ therefore $\frac{a}{r} ? 1$ . So $\frac{a^{2}}{r^{2}}$ is taken as negligible.

Therefore V = $\frac{2 q a^{2}}{4 π ε_{0} r^{3}}$

It can be inferred that V $\propto \frac{1}{r^{3}}$ . It is known that for a dipole V $\propto \frac{1}{r^{2}}$ and for monopole V $\propto \frac{1}{r}$

2.23 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 capacitors 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.

2.23 Requirements:

Capacitance, C = 2 $μ F$

Potential difference, V = 1 kV = 1000 V

Available:

Each capacitor, capacitance, $C_{1}$ = 1 $μ F$

Potential difference, $V_{1}$ = 400 V

Assumption:

A number of capacitors are connected in series and these series circuits are connected in parallel to each other. The potential difference across each row in parallel connection must be 1000 V and potential difference across each capacitor must be 400 V.

Hence, number of capacitors in each row is given as $\frac{1000}{250}$ = 2.5

So the number of capacitor in each row = 3

The equivalent capacitance in each row is given as, $\frac{1}{C_{R}}$ = $\frac{1}{1} + \frac{1}{1}$ + $\frac{1}{1}$ = 3, so $C_{R} = \frac{1}{3}$ $μ$ F

Let there be ‘n’ rows, each having 3 capacitors, which are connected in parallel

The total equivalent capacitance of the circuit is given as

$C_{e f f}$ = $\frac{1}{3}$ + $\frac{1}{3}$ + $\frac{1}{3}$ + $\frac{1}{3}$ + …….. upto n = $\frac{n}{3}$

It is given $C_{e f f}$ = 2. So n = 6

So the final circuit will be 3 capacitors in each row and 6 rows connected in parallel, Total capacitors required = 3 $\times 6 = 18$

2.24 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 realize 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 very minute separation between the conductors.]

2.24 Capacitance of the parallel capacitor, C = 2 F

Distance between two plates, d = 0.5 cm = 0.5 $\times 10^{- 2}$ m

Capacitance of a parallel plate capacitor is given by the relation,

C = $\frac{ε_{0} A}{d}$ , where $ε_{0} = P e r m i t t i v i t y o f f r e e s p a c e$ == 8.854 $\times 10^{- 12}$ $C^{2} N^{- 1}$ $m^{- 2}$

A = $\frac{C d}{ε_{0}}$ = $\frac{2 \times 0.5 \times 10^{- 2}}{8.854 \times 10^{- 12}}$ = 1129 $\times 10^{6}$ m = 1129 km

To avoid this situation, the capacitance of a capacitor is taken in $μ$ F.

2.25 Obtain the equivalent capacitance of the network in Fig. 2.33. For a 300 V supply, determine the charge and voltage across each capacitor.

2.25 Let C’ be the equivalent capacitance for capacitors $C_{2}$ and $C_{3}$ connected in series.

Hence, $\frac{1}{C'}$ = $\frac{1}{200}$ + $\frac{1}{200}$ . So C’ = 100 pF

Capacitors C’ and $C_{1}$ are in parallel, if the equivalent capacitance be C”, then

C” = C’ + $C_{1}$ = 100 + 100 = 200 pF

Now C” and $C_{4}$ are connected in series. If the total equivalent capacitance of the circuit be $C_{T o t a l}$ , then $\frac{1}{C_{T o t a l}}$ = $\frac{1}{C "}$ + $\frac{1}{C_{4}}$ = $\frac{1}{200}$ + $\frac{1}{100}$ , $C_{T o t a l}$ = $\frac{200}{3}$ pF = $\frac{200}{3} \times 10^{- 12}$ F

Let V” be the potential difference across C” and $V_{4}$ be the potential difference across $C_{4}$

Then, V” + $V_{4}$ = 300 V

Now, charge $Q_{4}$ on $C_{4}$ is given by $Q_{4}$ = $C_{T o t a l} \times V$ = $\frac{200}{3} \times 10^{- 12}$ $\times 300$ = 2 $\times 10^{- 8}$ C

$V_{4} = \frac{Q_{4}}{C_{4}}$ = $\frac{2 \times 10^{- 8}}{100 \times 10^{- 12}}$ = 200. So V” = 100 V

So potential difference across C’ and $C_{1}$ is V” = 100 V

Charge on $C_{1}$ is given by $Q_{1}$ = 100 $\times 10^{- 12} \times 100$ C = 1 $\times 10^{- 8}$ C

$C_{2}$ and $C_{3}$ are having same capacitance and have a potential difference of 100V together. Since $C_{2}$ and $C_{3}$ are in series, the potential difference is given by $V_{2}$ = $V_{3}$ = 50V

Hence the charge on $C_{2}$ is given by $Q_{2}$ = 200 $\times 10^{- 12} \times 50$ = $10^{- 8}$ C and the charge on $C_{3}$ is given by $Q_{3}$ = 200 $\times 10^{- 12} \times 50$ = $10^{- 8}$ C

Therefore, the equivalent capacitance of the circuit is $\frac{200}{3}$ pF and charge and voltage at all capacitance is given as

For $C_{1 :} Q_{1}$ = $10^{- 8}$ C, $V_{1}$ = 100 V

For $C_{2 :} Q_{2}$ = $10^{- 8}$ C, $V_{2}$ = 50 V

For $C_{3 :} Q_{3}$ = $10^{- 8}$ C, $V_{3}$ = 50 V

For $C_{4 :} Q_{4}$ = 2 $\times 10^{- 8}$ C, $V_{4}$ = 200 V

View other answers

2.26 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.

2.26 Area of the parallel plate capacitor, A = 90 ${c m}^{2}$ = 90 $\times 10^{- 4}$ $m^{2}$

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

Potential difference across plates, V = 400 V

Capacitance of the capacitor is given by, C = $\frac{ε_{0} A}{d},$

where $ε_{0} = P e r m i t t i v i t y o f f r e e s p a c e$ == 8.854 $\times 10^{- 12}$ $C^{2} N^{- 1}$ $m^{- 2}$

Electrostatic energy stored in the capacitor is given by the relation

$E_{1}$ = $\frac{1}{2}$ C $V^{2}$ = $\frac{1}{2}$ $\times \frac{ε_{0} A}{d} \times V^{2}$ = $\frac{1}{2}$ $\times \frac{8.854 \times 10^{- 12} \times 90 \times 10^{- 4}}{2.5 \times 10^{- 3}} \times 400^{2}$ = 2.55 $\times 10^{- 6}$ J

Volume of the given capacitor, V’ = A $\times d$ = 90 $\times 10^{- 4} \times$ 2.5 $\times 10^{- 3}$ $m^{3}$

= 2.25 $\times 10^{- 5}$ $m^{3}$

Energy stored is given by u = $\frac{E_{1}}{V ’}$ = $\frac{2.55 \times 10^{- 6}}{2.25 \times 10^{- 5}}$ = 0.113 J/ $m^{3}$

Also, u = $\frac{E_{1}}{V ’}$ = $\frac{\frac{1}{2} C V^{2}}{A d}$ = $\frac{\frac{1}{2} \times \frac{ε_{0} A}{d} \times V^{2}}{A d}$ = $\frac{1}{2} ε_{0} {(\frac{V}{d})}^{2}$ = $\frac{1}{2} ε_{0} E^{2}$ , where E = $\frac{V}{d}$ electric intensity

2.27 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 capacitor. How much electrostatic energy of the first capacitor is lost in the form of heat and electromagnetic radiation?

2.27 Capacitance of the charged capacitor, $C_{1}$ = 4 $μ F = 4 \times 10^{- 6}$ F

Supply voltage, $V_{1}$ = 200 V

Electrostatic energy stored in the $C_{1}$ capacitor, $E_{1}$ = $\frac{1}{2} C_{1} {V_{1}}^{2}$ = $\frac{1}{2} \times 4 \times 10^{- 6} \times 200^{2}$ J

= 0.08 J

Capacitance of the uncharged capacitor, $C_{2}$ = 2 $μ F$ = 2 $\times 10^{- 6}$ F

When $C_{2}$ is connected to $C_{1}$ , the potential acquired by $C_{2}$ be $V_{2}$

From the conservation of energy, the charge acquired by $C_{1}$ becomes the charge acquired by $C_{1}$ and $C_{2} .$

Hence $V_{2} \times (C_{1} + C_{2}) = C_{1} V_{1}$

or $V_{2}$ = $\frac{C_{1} V_{1}}{(C_{1} + C_{2})}$ = $\frac{4 \times 10^{- 6} \times 200}{(4 \times 10^{- 6} + 2 \times 10^{- 6})}$ = $\frac{400}{3}$ V

Electrostatic energy of the combination of two capacitors is given by

$E_{2}$ = $\frac{1}{2}$ ( $C_{1} + C_{2}) \times V_{2}^{2}$ = $\frac{1}{2} \times$ ( $4 \times 10^{- 6} +$ 2 $\times 10^{- 6}) \times {(\frac{400}{3})}^{2}$ = 5.33 $\times 10^{- 2}$ J

Hence the amount of electrostatic energy lost by capacitor $C_{1} = E_{1} - E_{2}$

= 0.08 - 5.33 $\times 10^{- 2}$ J = 0.0267 J = 2.67 $\times 10^{- 2}$ J

2.28 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 electric field between the plates. Explain the origin of the factor ½.

2.28 Let F be the force applied to separate the plates of a parallel plate capacitor and let $Δ x$ be the distance.

Hence work done by the force = F $Δ x$

The potential energy increase in the capacitor = uA $Δ x$ , where u = Energy density, A = area of each plate.

If d = distance between the plates and V = potential difference across the plates, the

Work done = increase in the potential energy.

Therefore,

F $Δ x$ = uA $Δ x$ or F = uA = ( $\frac{1}{2} ε_{0} E^{2}$ )A

Electric intensity is given by

E = $\frac{V}{d}$

F = ( $\frac{1}{2} ε_{0} E$ )EA= ( $\frac{1}{2} ε_{0} \frac{V}{d}$ )EA

Since Capacitance, C = $\frac{ε_{0} A}{d}$

F = ( $\frac{1}{2} C V$ E) = $\frac{1}{2}$ QE

2.29 A spherical capacitor consists of two concentric spherical conductors, held in position by suitable insulating supports (Fig. 2.34). Show that the capacitance of a spherical capacitor is given by

C = $\frac{4 π ε_{0} r_{1} r_{2}}{r_{1} - r_{2}}$

where $r_{1}$ and $r_{2}$ are the radii of outer and inner spheres, respectively.

2.29 Let F be the force applied to separate the plates of a parallel plate capacitor and let $Δ x$ be the distance.

Hence work done by the force = F $Δ x$

The potential energy increase in the capacitor = uA $Δ x$ , where u = Energy density, A = area of each plate.

If d = distance between the plates and V = potential difference across the plates, the

Work done = increase in the potential energy.

Therefore,

F $Δ x$ = uA $Δ x$ or F = uA = ( $\frac{1}{2} ε_{0} E^{2}$ )A

Electric intensity is given by

E = $\frac{V}{d}$

F = ( $\frac{1}{2} ε_{0} E$ )EA= ( $\frac{1}{2} ε_{0} \frac{V}{d}$ )EA

Since Capacitance, C = $\frac{ε_{0} A}{d}$

F = ( $\frac{1}{2} C V$ E) = $\frac{1}{2}$ QE

2.30 A spherical capacitor has an inner sphere of radius 12 cm and an outer sphere of radius 13 cm. The outer sphere is earthed and the inner sphere is given a charge of 2.5 μC. The space between the concentric spheres is filled with a liquid of dielectric constant 32.

(a) Determine the capacitance of the capacitor.

(b) What is the potential of the inner sphere?

(c) Compare the capacitance of this capacitor with that of an isolated sphere of radius 12 cm. Explain why the latter is much smaller.

2.30 Radius of the inner sphere, $r_{2}$ = 12 cm = 0.12 m

Radius of the outer sphere, $r_{1}$ = 13 cm = 0.13 m

Charges on the inner sphere, q= 2.5 $μ C = 2.5 \times 10^{- 6}$ C

Dielectric constant of the liquid, k = 32

Capacitance of the capacitor is given by the relation, C = $\frac{4 π ε_{0} {k r}_{1} r_{2}}{(r_{1} - r_{2})}$

$ε_{0}$ = permittivity of free space = 8.854 $\times 10^{- 12}$ $C^{2} N^{- 1}$ $m^{- 2}$

$C =$ $\frac{4 π \times 8.854 \times 10^{- 12} \times 32 \times 0.13 \times 0.12}{(0.13 - 0.12)}$ = 5.55 $\times 10^{- 9}$ F

Potential of the inner surface is given by

V = $\frac{q}{C}$ = $\frac{2.5 \times 10^{- 6}}{5.55 \times 10^{- 9}}$ = 450 V

Radius of the isolate sphere, r = 12 cm = 12 $\times 10^{- 2}$ m

Capacitance on the isolated sphere is given by C’ = 4 $π ε_{0}$ r

= 4 $\times π \times$ 8.854 $\times 10^{- 12} \times 12 \times 10^{- 2}$ F

= 1.34 $\times 10^{- 11}$ F

The capacitance of the isolated sphere is less in comparison to the concentric spheres. This is because the outer sphere of the concentric sphere is earthed, so the potential difference is less and the capacitance is more than the isolated sphere.

2.31 (a) Two 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 $Q_{1} Q_{2}$ /4 $π ε_{0} r^{2}$ , where r is the distance between their centres?

(b) If Coulomb’s law involved 1/ $r^{3}$ dependence (instead of 1/ $r^{2}$ ), would Gauss’s law be still true ?

(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?

(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?

(e) We know that electric field is discontinuous across the surface of a charged conductor. Is electric potential also discontinuous there?

(f) What meaning would you give to the capacitance of a single conductor?

(g) Guess a possible reason why water has a much greater dielectric constant (= 80) than say, mica (= 6).

2.31 The force between two conducting spheres is not exactly given by the expression $Q_{1} Q_{2}$ /4 $? ?_{0} r^{2}$ , because there is non-uniform charge distribution on the spheres.

Gauss's law will not be true, if Coulomb's law involved $\frac{1}{r^{3}}$ dependence, instead of $\frac{1}{r^{2}}$ , on r

Yes. If a small test charge is released at rest at a point in an electrostatic 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 the velocity.

Whenever the electron completes an orbit, either circular or elliptical, the work done by the field of a nucleus is zero.

No. Electric field is discontinuous across the surface of a charged conductor. However, electric potential is continuous.

The capacitance of a single conductor is considered as a parallel plate capacitor with one of its two plates at infinity.

Water has an unsymmetrical space as compared to mica. Since it has a permanent dipole moment, it has a greater dielectric constant than mica.

2.32 A cylindrical capacitor has two co-axial cylinders of length 15 cm and radii 1.5 cm and 1.4 cm. The outer cylinder is earthed and the inner cylinder is given a charge of 3.5 μC. 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).

2.32 Length of the co-axial cylinder, l = 15 cm = 0.15 m

Radius of the outer cylinder, $r_{1}$ = 1.5 cm = 0.015 m

Radius of the inner cylinder, $r_{2}$ = 1.4 cm = 0.014 m

Charge on the inner cylinder, q = 3.5 $μ C = 3.5 \times 10^{- 6}$ C

Capacitance of a co-axial cylinder of radii $r_{1}$ and $r_{2}$ is given by the relation

C = $\frac{2 π ε_{0} l}{\log_{e} ? \frac{r_{1}}{r_{2}}}$ , where $ε_{0}$ = permittivity of free space = 8.854 $\times 10^{- 12}$ $C^{2} N^{- 1}$ $m^{- 2}$

C = $\frac{2 π \times 8.854 \times 10^{- 12} \times 0.15}{\log_{e} ? \frac{0.015}{0.014}}$ = 1.21 $\times 10^{- 10}$ F

Potential difference of the inner cylinder is given by

V = $\frac{q}{C}$ = $\frac{3.5 \times 10^{- 6}}{1.21 \times 10^{- 10}}$ = 2.893 $\times 10^{4}$ V

2.33 A parallel plate capacitor is to be designed with a voltage rating 1 kV, using a material of dielectric constant 3 and dielectric strength about 10⁷ Vm–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 50 pF?

2.33 Potential rating of the capacitor, V = 1 kV = 1000 V

Dielectric constant of a material, $?_{r}$ = 3

Dielectric strength = $10^{7}$ V/m

For safety, the field intensity should not cross 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 = 50 pF = 50 $\times 10^{- 12}$ F

Distance between the plates, d is given by d = $\frac{V}{E}$ = $\frac{1000}{10^{6}}$ m = $10^{- 3}$ m

Capacitance is given by the relation

C = $\frac{ε_{0} ?_{r} A}{d}$ , where $ε_{0}$ = permittivity of free space = 8.854 $\times 10^{- 12}$ $C^{2} N^{- 1}$ $m^{- 2}$

50 $\times 10^{- 12}$ = $\frac{8.854 \times 10^{- 12} \times 3 \times A}{10^{- 3}}$

A = 1.88 $\times 10^{- 3}$ $m^{2}$ = 18.8 ${c m}^{2}$

Hence, the area of each plate is about 19 ${c m}^{2}$

2.34 Describe schematically the equipotential surfaces corresponding to

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

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

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

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

2.34 Equidistant planes parallel to the x-y plane are the equipotential surfaces.

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

Concentric spheres centered at the origin are equipotential surfaces.

A periodically varying 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.

2.35 A small sphere of radius r₁ and charge q₁ is enclosed by a spherical shell of radius r₂ and charge q₂. Show that if q₁ 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 q₂ on the shell is.

2.35 According to Gauss's law, the electric field between a sphere and a shell is determined by the charge $q_{1}$ on a small sphere. Hence, the potential difference, V, between the sphere and the shell is independent of charge $q_{2}$ . For positive charge $q_{1}$ , potential difference V s always positive.

2.36 Answer the following:

(a) The top of the atmosphere is at about 400 kV 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 100 Vm–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!)

(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 1m2. Will he get an electric shock if he touches the metal sheet next morning?

(c) The discharging current in the atmosphere due to the small conductivity of air is known to be 1800 A 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?

(d) What are the forms of energy into which the electrical energy of the atmosphere is dissipated during a lightning?

(Hint: The earth has an electric field of about 100 Vm–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 50 km (beyond which it is good conductor), about + 1800 C 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.)

2.36 We do not get an electric shock as we step out of our house because the original equipotential surfaces of open air changes, keeping our body and the ground at the same potential.

Yes, the person will get an electric shock if he touches the metal slab next morning. The steady discharging current in the atmosphere charges up the aluminium sheet. As a result, its voltage rises gradually. The raise in the voltage depends on the capacitance of the capacitor formed by the aluminium slab and the ground.

The occurrence of thunderstorm and lightning charges the atmosphere continuously. Hence, even with the presence of discharging current of 1800 A, the atmosphere is not discharged completely. The two opposing currents are in equilibrium and the atmosphere remains electrically neutral.

During lightning and thunderstorm, light energy, heat energy and sound enegy are dissipated in the atmosphere.

Complete Study Material - Electrostatic Potential and Capacitance

Access complete study material, including topic-wise notes, revision notes, exercise solutions, and more for Class 12 Physics Chapter 2. Find the links in the table below.

Related Class 12 Chapter 2 Study Material
Class 12 Physics Chapter 2 NCERT Exemplar Solutions
Electrostatic Potential and Capacitance NCERT Notes
Electrostatic Potential and Capacitance Class 12 Formulas
Class 12 Electrostatic Potential and Capacitance Quick Revision Notes

Key Topics in Class 12 Electrostatic Potential and Capacitance

Here are all the topics covered in Class 12 Chapter 2 of Physics. These topics are really important for the CBSE Board exam. Frequently, numerical and theoretical questions related to capacitors and energy stored, potential difference, and cells in JEE Mains and NEET.

Exercise	Topics Covered
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
2.7	Potential Energy Of A System Of Charges
2.8	Potential Energy In An External Field
2.9	Electrostatics Of Conductors

Chapter 2 Electrostatic Potential and Capacitance Important Formulas

You can read all the important formulas of Electrostatic Potential and Capacitance in the table below.

Formula Name	Expression
Electrostatic Potential (V)	V = W / q
Potential due to a Point Charge	V = (1 / (4 * π * ε₀)) * (q / r)
Potential due to Electric Dipole at a Point (on Axial Line)	V = (1 / (4 * π * ε₀)) * (p * cosθ / r²)
Potential Difference Between Two Points (A to B)	V_B - V_A = -∫ E · dl
Relation Between Electric Field and Potential	E = -∇V
Work Done Along an Equipotential Surface	W = 0
Capacitance of a Capacitor	C = q / V
Capacitance of a Parallel Plate Capacitor	C = ε₀ * A / d
Capacitance with Dielectric Between Plates	C = K * ε₀ * A / d
Capacitance in Series	1 / C_eq = 1 / C₁ + 1 / C₂ + ... + 1 / Cₙ
Capacitance in Parallel	C_eq = C₁ + C₂ + ... + Cₙ
Electric Potential Energy Stored in a Capacitor	U = 1/2 * C * V² = 1/2 * q² / C = 1/2 * q * V
Energy Density of Electric Field	u = 1/2 * ε₀ * E²
Polarization (P)	P = p_m / V
Electric Field in a Dielectric	E = E₀ / K
Induced Surface Charge Density	σ_ind = P · n̂

Chapter 2 Class 12 Physics Weightage

You can check the weightage in different competitive exams of this chapter below.

Exam	Weightage
CBSE Boards	5 - 8 Marks
JEE Mains	4 - 8 Marks
NEET	2 - 5%
KCET	5 - 8%

How to Use NCERT Solutions for Electrostatic Potential And Capacitance

Having the NCERT Solutions by your side and using these solutions are two different things. You can use these tips to utilise solutions in a better way.

Read the NCERT Textbook and solve the NCERT Examples given in the box in the book.
Read our topic-wise notes and build your concept before going through the NCERT Exercise.
Use the chapter-wise formulas list provided by Shiksha to memorise all the formulas.
Now, solve your Chapter 2 NCERT Exercise of Class 12th Physics and take help of solutions to find mistakes and correct methods.
Practice your answer-writing skills to build confidence for Class 12th Board exams.

NCERT Physics Chapter 2 Electrostatic Potential and Capacitance – FAQs

The following are the frequently asked questions on Chapter 2, Electrostatic Potential and Capacitance:

Commonly asked questions

What is electric potential and capacitance?

The ability to do work on a charge is called the electric potential and the ability to store charge is termed as the capacitance.

What is the weightage of electric potential class 12 in NEET exam?

In NEET Physics exam, the chapter electric potential and capacitance carries a weightage of 2% to 5%, which means you can expect one question out of 45 questions.

What is electric potential and capacitance class 12 weightage in JEE Main exam?

In JEE Main Physics, electric potential and capacitance chapter has a weightage of 3% to 6%. You can expect around 1 or 2 questions from this chapter which carries 4 to 8 marks.

Explain the SI Unit of Capacitance.

It is the farad (F). The name came from Michael Faraday. SI Unit of Capacitance means the ability of a system to store an electric charge. One farad is the capacitance of a device that needs one coulomb of charge to provide one volt potential difference across it. Mathematically, it is represented by - 1 Farad = 1 Coulomb/Volt.

Is electric potential and capacitance class 12 a hard chapter?

This chapter covers the concepts of potential, capacitors, and potential energy. It is considered as one of the easy chapter of the class 12 Physics.

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Class 12 Physics Chapter 2 Electrostatic Potential And Capacitance Solutions: Free PDF and Important Formulas

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Key Topics in Class 12 Electrostatic Potential and Capacitance

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