Electrostatic Potential and Capacitance: Overview, Questions, Preparation

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 =  $\frac{1}{4\pi {\epsilon }_0}\times \frac{q_1}{r}$  $+$  $\frac{1}{4\pi {\epsilon }_0}\times \frac{q_2}{(d-r)}$ …………………………..(1)

Where  ${\epsilon }_0$ = permittivity of free space

For V = 0, the equation (1) becomes

$\frac{1}{4\pi {\epsilon }_0}\times \frac{q_1}{r}=-\frac{1}{4\pi {\epsilon }_0}\times \frac{q_2}{(d-r)}$ or  $\frac{q_1}{r}$ =  $-\frac{q_2}{(d-r)}$ or  $\frac{5\mathrm{}\times {10}^{-8}}{r}$ =  $-\frac{-3\mathrm{}\times {10}^{-8}}{(d-r)}$

$\frac{5\mathrm{}\times {10}^{-8}}{r}$ =  $\frac{3\mathrm{}\times {10}^{-8}}{(d-r)}$ or  $\frac{r}{(d-r)}$ =  $\frac{5\mathrm{}\times {10}^{-8}}{3\mathrm{}\times {10}^{-8}}$ 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 =  $\frac{1}{4\pi {\epsilon }_0}\times \frac{q_1}{s}$  $+$  $\frac{1}{4\pi {\epsilon }_0}\times \frac{q_2}{(s-d)}$ ……………………………(2)

For V = 0, we get

$\frac{1}{4\pi {\epsilon }_0}\times \frac{q_1}{s}=-\frac{1}{4\pi {\epsilon }_0}\times \frac{q_2}{(s-d)}$ or  $\frac{q_1}{s}=-\frac{q_2}{(s-d)}$ or  $\frac{5\mathrm{}\times {10}^{-8}}{s}=-\frac{-3\mathrm{}\times {10}^{-8}}{(s-d)}$

$\frac{5}{s}$ =  $\frac{3}{s-0.16}$ 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  $\mu C=5\times {10}^{-6}$ 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 =  $\frac{1}{4\pi {\epsilon }_0}\times \frac{6q_1}{d}$ , where  ${\epsilon }_0$ = permittivity of free space = 8.854  $\times {10}^{-12}$  $C^2{N}^{-1}$  $m^{-2}$

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

(c) at a point 18 cm from the centre of 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 =  $\frac{1}{4\pi {\epsilon }_0}\times \frac{q}{r^2}$ where  ${\epsilon }_0$ = permittivity of free space = 8.854  $\times {10}^{-12}$  $C^2{N}^{-1}$  $m^{-2}$

E =  $\frac{1}{4\times \pi \times 8.854\mathrm{}\times {10}^{-12}}\times \frac{1.6\mathrm{}\times {10}^{-7}}{{0.12\mathrm{}}^2}$ = 99863.8 N  $C^{-1}$  $\cong {10}^5$ 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 =  $\frac{1}{4\times \pi \times 8.854\mathrm{}\times {10}^{-12}}\times \frac{1.6\mathrm{}\times {10}^{-7}}{{0.18\mathrm{}}^2}$ = 4.438  $\times {10}^4$ 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?

Ans.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 =  $\frac{k{\epsilon }_{0}A}{d}$ =  $\frac{{\epsilon }_{0}A}{d}$ , since k = 1……………………..(1)

Where A = area of each plate

${\epsilon }_0$ = 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,  $C_1$ =  $\frac{6{\epsilon }_{0}A}{d/2}$ =  $\frac{12{\epsilon }_{0}A}{d}$ ……………(2)

From equation (1) and (2) we get

$C_1=12C$ = 12  $\times 8$ pF = 96 pF

 

Q.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?

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

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

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

Hence,  $C_{eq}$ = 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

 

Q.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.

Ans.2.7 Let the three capacitors be

$C_1$ = 2 pF,  $C_2$ = 3 pF and  $C_3$ = 4 pF

The equivalent capacitance,  $C_{eq}$ is given by

$C_{eq}=C_1+C_2+C_3$ = 2 + 3 + 4 = 9 pF

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

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

Hence for  $C_1$ ,  $q_1$ = 100  $\times 2pC$ = 200 pC = 2  $\times {10}^{-10}$ C,

for  $C_2$ ,  $q_2$ = 100  $\times 3pC$ = 300 pC = 3  $\times {10}^{-10}$ C,

for  $C_3$ ,  $q_3$ = 100  $\times 4pC$ = 400 pC = 4  $\times {10}^{-10}$ C

 

Q.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?

Ans.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 =  $\frac{k{\epsilon }_{0}A}{d}$

In case of air, dielectric constant k = 1 and

${\epsilon }_0$ = permittivity of free space = 8.854  $\times {10}^{-12}$  $C^2{N}^{-1}$  $m^{-2}$

Hence C =  $\frac{8.854\mathrm{}\times {10}^{-12}\times 6\mathrm{}\times {10}^{-3}}{3\mathrm{}\times {10}^{-3}}$ 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

 

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.

Ans.2.9 Dielectric constant of the mica sheet, k =6

While the voltage supply remained connected :

V = 100 V

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

New capacitance,  $C_1$ = kC = 6  $\times$ 17.708  $\times {10}^{-12}F$ = 106.25  $\times {10}^{-12}$ F

= 106.25 pF

New charge,  $q_1$ =  $C_{1}V$ = 106.25  $\times {10}^{-12}\times 100$ C = 10.62  $\times {10}^{-9}$ C

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

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

Potential across plates,  $V_1$ =  $\frac{q}{C_1}$ =  $\frac{1.7708\mathrm{}\times {10}^{-9}}{106.25\mathrm{}\times {10}^{-12}}$ = 16.66 V

 

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

Ans.2.10 Capacitance C = 12 pF = 12  $\times {10}^{-12}$ F

Potential difference, V = 50V

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

E =  $\frac{1}{2}$ C  $V^2$ =  $\frac{1}{2}\times$ 12  $\times {10}^{-12}\times {\left(50\right)}^2$ J = 1.5  $\times {10}^{-8}$ 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?

Ans.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  $C_{eq}$ is given by

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

$C_{eq}=300pF$

New electrostatic energy,  $E_2$ =  $\frac{1}{2}{C}_{eq}{V}^2$ =  $\frac{1}{2}\times 300\times {10}^{-12}\times {\left(200\right)}^2$ = 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

 

Q.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).

Ans.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,  $V_1$ =  $\frac{q}{4\pi {\epsilon }_0{d}_1}$

Potential at point Q,  $V_2$ =  $\frac{q}{4\pi {\epsilon }_0{d}_2}$

${\epsilon }_0$ = permittivity of free space = 8.854  $\times {10}^{-12}$  $C^2{N}^{-1}$  $m^{-2}$

Work done, W =  $q_1$  $(V_2-V_1)$ =  $\frac{q{q}_1}{4\pi {\epsilon }_0}(\frac{1}{d_2}\mathrm{}-\mathrm{}\frac{1}{d_1})\mathrm{}$ =  $\frac{8\mathrm{}\times {10}^{-3}\times -2\mathrm{}\times {10}^{-9}}{4\times \pi \times 8.854\mathrm{}\times {10}^{-12}}(\frac{1}{0.04}\mathrm{}-\mathrm{}\frac{1}{0.03})\mathrm{}$

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

 

Q.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.

Ans.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  $\surd 2$

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

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

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

V =  $\frac{8q}{4\pi {\epsilon }_{0}r}$ =  $\frac{8q}{4\pi {\epsilon }_0\frac{\surd 3}{2}b}$ =  $\frac{4q}{\surd 3\pi {\epsilon }_{0}b}$

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

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.

 

Q.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.

Ans.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  $\mu C$ = 1.5  $\times {10}^{-6}C$ and

at point B,  $q_2$ = 2.5  $\mu C=2.5\times {10}^{-6}C$

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

$=\frac{q_1}{4\pi {\epsilon }_0\frac{d}{2}}$ +  $\frac{q_2}{4\pi {\epsilon }_0\frac{d}{2}}$ =  $\frac{1}{4\pi {\epsilon }_0\frac{d}{2}}$ (  $q_1+q_2)$

${\epsilon }_0$ = permittivity of free space = 8.854  $\times {10}^{-12}$  $C^2{N}^{-1}$  $m^{-2}$

$V_1=$  $\frac{1}{4\pi \times 8.854\mathrm{}\times {10}^{-12}\times \frac{0.3}{2}}$ (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  $q_1$

$=$  $\frac{q_2}{4\pi {\epsilon }_0({\frac{d}{2})}^2}$  $-$  $\frac{q_1}{4\pi {\epsilon }_0({\frac{d}{2})}^2}$

$=$  $\frac{1}{4\pi {\epsilon }_0({\frac{d}{2})}^2}(q_2-q_1)$ =  $\frac{1}{4\pi \times 8.854\mathrm{}\times {10}^{-12}\times ({\frac{0.3}{2})}^2}(2.5\times {10}^{-6}-1.5\mathrm{}\times {10}^{-6})$

= 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

$=\frac{q_1}{4\pi {\epsilon }_{0}AZ}$ +  $\frac{q_2}{4\pi {\epsilon }_{0}BZ}$ =  $\frac{1}{4\pi {\epsilon }_{0}0.18}$ (  $q_1+q_2)$

${\epsilon }_0$ = permittivity of free space = 8.854  $\times {10}^{-12}$  $C^2{N}^{-1}$  $m^{-2}$

$V_2=$  $\frac{1}{4\pi \times 8.854\mathrm{}\times {10}^{-12}\times 0.18}$ (1.5  $\times {10}^{-6}+2.5\times {10}^{-6})$ = 2.0  $\times {10}^5$ V

Electric field due to charge  $q_{1}atZ,E_A$

$E_A=\frac{q_1}{4\pi {\epsilon }_0({AZ)}^2}$ =  $\frac{1.5\mathrm{}\times {10}^{-6}}{4\times \pi \times 8.854\mathrm{}\times {10}^{-12}\times ({0.18)}^2}$ = 0.416  $\times {10}^6$ V/m

Electric field due to charge  $q_{2}atZ,E_B$

$E_B=\frac{q_2}{4\pi {\epsilon }_0({BZ)}^2}$ =  $\frac{2.5\mathrm{}\times {10}^{-6}}{4\times \pi \times 8.854\mathrm{}\times {10}^{-12}\times ({0.18)}^2}$ = 0.693  $\times {10}^6$ V/m

The resultant intensity at Z

$E_2$ =  $\sqrt{{E_A}^2+{E_B}^2+2E_A{E}_{B}cos2\theta }$ where 2  $\theta$ is the  $\angle AZB$

From the figure, we get  $\cos \theta$ =  $\frac{ZO}{ZB}$ =  $\frac{0.1}{0.18}$

Hence  $\theta$ = 56.25  $°$

Therefore,  $E_2$ =  $E_2=\sqrt{({0.416\mathrm{}\times {10}^6)}^2+{(0.693\mathrm{}\times {10}^6)}^2+2{\left(0.416\mathrm{}\times {10}^6\right)\times (0.693\mathrm{}\times {10}^6)}_{}cos2\times 56.25\mathrm{}°}$ = 6.57  $\times {10}^5$ V/m

 

Q.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.

Ans.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,

${\sigma }_1=\frac{Totalcharge}{Innersurfacearea}=\frac{-q}{4\pi r_1^2}$ …………(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

${\sigma }_2=\frac{Totalcharge}{Outersurfacearea}=\frac{Q+q}{4\pi r_2^2}$ …………(2)

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

 

Q.2.16 (a) Show that the normal component of electrostatic field has a discontinuity from one side of a charged surface to another given by

$({\mathit{E}}_2$ -  ${\mathit{E}}_1).\stackrel{̂}{\mathit{n}}=\frac{\mathit{\sigma }}{{\mathit{\epsilon }}_0}$

Where  $\stackrel{̂}{\mathit{n}}$ is a unit vector normal to the surface at a point and s is the surface charge density at that point. (The direction of  $\stackrel{̂}{\mathit{n}}$ is from side 1 to side 2.) Hence, show that just outside a conductor, the electric field is  $\mathit{\sigma }\stackrel{̂}{\mathit{n}}$ /  ${\mathit{\epsilon }}_0$ .

(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.]

Ans.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,

$\overrightarrow{E_1}$ =  $\frac{\sigma }{2{\epsilon }_0}\stackrel{̂}{n}$ ………….(i)

Where,

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

$\sigma$ = surface charge density at that point

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

$\overrightarrow{E_2}$ =  $\frac{\sigma }{2{\epsilon }_0}\stackrel{̂}{n}$ ………….(ii)

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

$\overrightarrow{E_2}-\overrightarrow{E_1}=\frac{\sigma }{2{\epsilon }_0}\stackrel{̂}{n}+\frac{\sigma }{2{\epsilon }_0}\stackrel{̂}{n}=\frac{\sigma }{{\epsilon }_0}\stackrel{̂}{n}$ ……(iii)

$\mathrm{S}\mathrm{i}\mathrm{n}\mathrm{c}\mathrm{e}\mathrm{}\mathrm{i}\mathrm{n}\mathrm{s}\mathrm{i}\mathrm{d}\mathrm{e}\mathrm{}\mathrm{a}\mathrm{}\mathrm{c}\mathrm{l}\mathrm{o}\mathrm{s}\mathrm{e}\mathrm{d}\mathrm{}\mathrm{c}\mathrm{o}\mathrm{n}\mathrm{d}\mathrm{u}\mathrm{c}\mathrm{t}\mathrm{o}\mathrm{r},\mathrm{}\overrightarrow{{\mathrm{E}}_1}=0,$

$\vec{E}$ =  $\overrightarrow{E_2}$ =  $\frac{\sigma }{{\epsilon }_0}\stackrel{̂}{n}$

Therefore, the electric field just outside the conductor is  $\frac{\sigma }{{\epsilon }_0}\stackrel{̂}{n}$

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.

 

Q.2.17 A long charged cylinder of linear charged density  $\mathit{\lambda }\mathit{}$ is surrounded by a hollow co-axial conducting cylinder. What is the electric field in the space between the two cylinders?

Ans.2.17

 

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

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

$\phi =E(2\pi$ dL) =  $\frac{q}{{\epsilon }_0}$ , where

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

${\mathrm{\epsilon }}_0=\mathrm{P}\mathrm{e}\mathrm{r}\mathrm{m}\mathrm{i}\mathrm{t}\mathrm{t}\mathrm{i}\mathrm{v}\mathrm{i}\mathrm{t}\mathrm{y}\mathrm{}\mathrm{o}\mathrm{f}\mathrm{}\mathrm{f}\mathrm{r}\mathrm{e}\mathrm{e}\mathrm{}\mathrm{s}\mathrm{p}\mathrm{a}\mathrm{c}\mathrm{e}$

Then,  $\phi =E(2\pi$ dL) =  $\frac{q}{{\epsilon }_0}$ =  $\frac{\lambda L}{{\epsilon }_0}$

E =  $\frac{\lambda }{{2\pi d\epsilon }_0}$

Therefore, the electric field in the space between the two cylinders is  $\frac{\lambda }{{2\pi d\epsilon }_0}.$

 

Q.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?

Ans.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 -  $\frac{q_1{q}_2}{4\pi {\epsilon }_{0}d}$ , where  ${\mathrm{\epsilon }}_0=\mathrm{P}\mathrm{e}\mathrm{r}\mathrm{m}\mathrm{i}\mathrm{t}\mathrm{t}\mathrm{i}\mathrm{v}\mathrm{i}\mathrm{t}\mathrm{y}\mathrm{}\mathrm{o}\mathrm{f}\mathrm{}\mathrm{f}\mathrm{r}\mathrm{e}\mathrm{e}\mathrm{}\mathrm{s}\mathrm{p}\mathrm{a}\mathrm{c}\mathrm{e}$ == 8.854  $\times {10}^{-12}$  $C^2{N}^{-1}$  $m^{-2}$

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

=  $\frac{q_1{q}_2}{4\pi {\epsilon }_{0}d}$  $\times \frac{1}{1.6\mathrm{}\times {10}^{-19}}$ - 27.13 eV =  $\frac{{(1.6\mathrm{}\times {10}^{-19})}^2}{4\pi \times 8.854\mathrm{}\times {10}^{-12}\times 1.06\mathrm{}\times {10}^{-10}}\times \frac{1}{1.6\mathrm{}\times {10}^{-19}}-$ 27.13 = 13.56 -27.13 eV= -13.56 eV

 

Q.2.19 If one of the two electrons of a  ${\mathit{H}}_2$ molecule is removed, we get a hydrogen molecular ion  ${\mathit{H}}_2^{+}$ . In the ground state of an  ${\mathit{H}}_2^{+}$ , 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.

$\mathit{}$ Ans.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 =  $\frac{q_1{q}_2}{4\pi {\epsilon }_0{d}_1}$ +  $\frac{q_2{q}_3}{4\pi {\epsilon }_0{d}_3}$ +  $\frac{q_3{q}_1}{4\pi {\epsilon }_0{d}_2}$ ,

Where  ${\mathrm{\epsilon }}_0=\mathrm{P}\mathrm{e}\mathrm{r}\mathrm{m}\mathrm{i}\mathrm{t}\mathrm{t}\mathrm{i}\mathrm{v}\mathrm{i}\mathrm{t}\mathrm{y}\mathrm{}\mathrm{o}\mathrm{f}\mathrm{}\mathrm{f}\mathrm{r}\mathrm{e}\mathrm{e}\mathrm{}\mathrm{s}\mathrm{p}\mathrm{a}\mathrm{c}\mathrm{e}$ = 8.854  $\times {10}^{-12}$  $C^2{N}^{-1}$  $m^{-2}$

V =  $\frac{1}{4\pi {\epsilon }_0}$ (  $\frac{q_1{q}_2}{d_1}$ +  $\frac{q_2{q}_3}{d_2}$ +  $\frac{q_3{q}_1}{d_2}$ )

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

= 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

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

= - 19.17 eV

 

Q.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.

Ans.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 =  $E_B$

$E_A\mathrm{i}\mathrm{s}\mathrm{}\mathrm{g}\mathrm{i}\mathrm{v}\mathrm{e}\mathrm{n}\mathrm{}\mathrm{b}\mathrm{y},E_A=$  $\frac{Q_A}{4\pi {\epsilon }_0{a}^2}$ and  $E_B\mathrm{i}\mathrm{s}\mathrm{}\mathrm{g}\mathrm{i}\mathrm{v}\mathrm{e}\mathrm{n}\mathrm{}\mathrm{b}\mathrm{y},E_B=$  $\frac{Q_B}{4\pi {\epsilon }_0{b}^2}$

$\frac{E_A}{E_B}$ =  $\frac{Q_A}{Q_B}\times \frac{b^2}{a^2}$ ………..(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

$\frac{E_A}{E_B}$ =  $\frac{a}{b}\times \frac{b^2}{a^2}$ =  $\frac{b}{a}$

 

Q.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?

Ans.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 =  $\frac{1}{4\pi {\epsilon }_0}$ (  $\frac{q}{z-a})+\frac{1}{4\pi {\epsilon }_0}$ (  $-\frac{q}{z+a})$ =  $\frac{q(z+a-z+a)}{4\pi {\epsilon }_0(z^2-a^2)}$ =  $\frac{2qa}{4\pi {\epsilon }_0(z^2-a^2)}$ =  $\frac{p}{4\pi {\epsilon }_0(z^2-a^2)}$

Where  ${\epsilon }_0$ = 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

$V_1$ =  $\frac{-q}{4\pi {\epsilon }_0}\frac{1}{\sqrt{{(5-0)}^2+{(-a)}^2}}+\frac{q}{4\pi {\epsilon }_0}\frac{1}{\sqrt{{(5-0)}^2+{\left(a\right)}^2}}$

= 0

Electrostatic potential (  $V_2)$ at point (-7,0,0) is given by

$V_2$ =  $\frac{-q}{4\pi {\epsilon }_0}\frac{1}{\sqrt{{(-7)}^2+{(-a)}^2}}+\frac{q}{4\pi {\epsilon }_0}\frac{1}{\sqrt{{(-7)}^2+{\left(a\right)}^2}}$

= 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.

 

Q.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).

 

Ans.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 =  $\frac{1}{4\pi {\epsilon }_0}[\frac{q}{XP}$ -  $\frac{2q}{YP}$ +  $\frac{q}{ZP}$ ]

=  $\frac{1}{4\pi {\epsilon }_0}[\frac{q}{r+a}$ -  $\frac{2q}{r}$ +  $\frac{q}{r-a}$ ]

=  $\frac{q}{4\pi {\epsilon }_0}[\frac{r\left(r-a\right)-2\left(r-a\right)\left(r+a\right)+r(r+a)}{r\left(r-a\right)(r+a)}$ ]

=  $\frac{q}{4\pi {\epsilon }_0}[\frac{r^2-ra-2r^2+2a^2+r^2+ra}{r\left(r-a\right)(r+a)}$ ]

=  $\frac{q}{4\pi {\epsilon }_0}[\frac{2a^2}{r(r^2-a^2)}$ ]

=  $\frac{2q{a}^2}{4\pi {\epsilon }_0{r}^3(1-\frac{a^2}{r^2})}$

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

Therefore V =  $\frac{2q{a}^2}{4\pi {\epsilon }_0{r}^3}$

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  $\propto \frac{1}{r}$

 

Q.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.

Ans.2.23 Requirements:

Capacitance, C = 2  $\mu F$

Potential difference, V = 1 kV = 1000 V

Available:

Each capacitor, capacitance,  $C_1$ = 1  $\mu F$

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  $C_R=\frac{1}{3}$  $\mu$ 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

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

It is given  $C_{eff}$ = 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  $\times 6=18$

 

Q.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.]

Ans.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 =  $\frac{{\epsilon }_{0}A}{d}$ , where  ${\mathrm{\epsilon }}_0=\mathrm{P}\mathrm{e}\mathrm{r}\mathrm{m}\mathrm{i}\mathrm{t}\mathrm{t}\mathrm{i}\mathrm{v}\mathrm{i}\mathrm{t}\mathrm{y}\mathrm{}\mathrm{o}\mathrm{f}\mathrm{}\mathrm{f}\mathrm{r}\mathrm{e}\mathrm{e}\mathrm{}\mathrm{s}\mathrm{p}\mathrm{a}\mathrm{c}\mathrm{e}$ == 8.854  $\times {10}^{-12}$  $C^2{N}^{-1}$  $m^{-2}$

A =  $\frac{Cd}{{\mathrm{\epsilon }}_0}$ =  $\frac{2\times 0.5\mathrm{}\times {10}^{-2}}{8.854\mathrm{}\times {10}^{-12}}$ = 1129  $\times {10}^6$ m = 1129 km

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

 

Q.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.

 

Let C’ be the equivalent capacitance for capacitors  $C_2$ and  $C_3$ connected in series.

Hence,  $\frac{1}{C\text{'}}$ =  $\frac{1}{200}$ +  $\frac{1}{200}$ . 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  $C_{Total}$ , then  $\frac{1}{C_{Total}}$ =  $\frac{1}{C"}$ +  $\frac{1}{C_4}$ =  $\frac{1}{200}$ +  $\frac{1}{100}$ ,  $C_{Total}$ =  $\frac{200}{3}$ 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  $Q_4$ =  $C_{Total}\times V$ =  $\frac{200}{3}\times {10}^{-12}$  $\times 300$ = 2  $\times {10}^{-8}$ C

$V_4=\frac{Q_4}{C_4}$ =  $\frac{2\mathrm{}\times {10}^{-8}}{100\times {10}^{-12}}$ = 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

 

Q.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.

Ans.2.26 Area of the parallel plate capacitor, A = 90  ${cm}^2$ = 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 =  $\frac{{\epsilon }_{0}A}{d},$

where  ${\mathrm{\epsilon }}_0=\mathrm{P}\mathrm{e}\mathrm{r}\mathrm{m}\mathrm{i}\mathrm{t}\mathrm{t}\mathrm{i}\mathrm{v}\mathrm{i}\mathrm{t}\mathrm{y}\mathrm{}\mathrm{o}\mathrm{f}\mathrm{}\mathrm{f}\mathrm{r}\mathrm{e}\mathrm{e}\mathrm{}\mathrm{s}\mathrm{p}\mathrm{a}\mathrm{c}\mathrm{e}$ == 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  $V^2$ =  $\frac{1}{2}$  $\times \frac{{\epsilon }_{0}A}{d}\times V^2$ =  $\frac{1}{2}$  $\times \frac{8.854\mathrm{}\times {10}^{-12}\times 90\mathrm{}\times {10}^{-4}}{2.5\mathrm{}\times {10}^{-3}}\times {400}^2$ = 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 =  $\frac{E_1}{\mathrm{V}\mathrm{’}}$ =  $\frac{2.55\mathrm{}\times {10}^{-6}}{2.25\mathrm{}\times {10}^{-5}}$ = 0.113 J/  $m^3$

Also, u =  $\frac{E_1}{\mathrm{V}\mathrm{’}}$ =  $\frac{\frac{1}{2}C{V}^2}{Ad}$ =  $\frac{\frac{1}{2}\mathrm{}\times \frac{{\epsilon }_{0}A}{d}\times V^2}{Ad}$ =  $\frac{1}{2}{\epsilon }_0{\left(\frac{V}{d}\right)}^2$ =  $\frac{1}{2}{\epsilon }_0{\mathrm{E}}^2$ , where E =  $\frac{V}{d}$ electric intensity

 

Q.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?

Ans.2.27 Capacitance of the charged capacitor,  $C_1$ = 4  $\mu F=4\times {10}^{-6}$ 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  $\mu F$ = 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  $V_2\times \left(C_1+C_2\right)=C_1{V}_1$

or  $V_2$ =  $\frac{C_1{V}_1}{\left(C_1+C_2\right)}$ =  $\frac{4\times {10}^{-6}\times 200}{(4\times {10}^{-6}+2\mathrm{}\times {10}^{-6})}$ =  $\frac{400}{3}$ 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  $\times {10}^{-6})\times {\left(\frac{400}{3}\right)}^2$ = 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

 

Q.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.

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

Hence work done by the force = F  $\Delta x$

The potential energy increase in the capacitor = uA  $\Delta x$ , 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  $\Delta x$ = uA  $\Delta x$ or F = uA = (  $\frac{1}{2}{\epsilon }_0{E}^2$ )A

Electric intensity is given by

E =  $\frac{V}{d}$

F = (  $\frac{1}{2}{\epsilon }_{0}E$ )EA= (  $\frac{1}{2}{\epsilon }_0\frac{V}{d}$ )EA

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

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

 

Q.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 =  $\frac{4\mathit{\pi }{\mathit{\epsilon }}_0{\mathit{r}}_1{\mathit{r}}_2}{{\mathit{r}}_1-{\mathit{r}}_2}$

 

where  ${\mathit{r}}_1$ and  ${\mathit{r}}_2\mathit{}$ are the radii of outer and inner spheres, respectively.

Ans.2.29 Radius of the outer shell =  $r_1$ , radius of the inner shell =  $r_2$

The charge at the inner surface of the outer shell = +Q

The outer surface of the inner shell has induced charge –Q

Potential difference between the two shells is given by,

V =  $\frac{Q}{4\pi {\epsilon }_0{r}_2}$  $-\frac{Q}{4\pi {\epsilon }_0{r}_1}$ where  ${\epsilon }_0$ is the permittivity of free space

V =  $\frac{Q}{4\pi {\epsilon }_0}$ (  $\frac{1}{r_2}$ -  $\frac{1}{r_1})$ =  $\frac{Q(r_1-r_2)}{4\pi {\epsilon }_0{r}_1{r}_2}$

Capacitance of the given system C =  $\frac{Q}{V}$ =  $\frac{4\pi {\epsilon }_0{r}_1{r}_2}{(r_1-r_2)}$

 

Q.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.

Ans.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  $\mu C=2.5\times {10}^{-6}$ C

Dielectric constant of the liquid, k = 32

Capacitance of the capacitor is given by the relation, C =  $\frac{4\pi {\epsilon }_0{kr}_1{r}_2}{(r_1-r_2)}$

${\epsilon }_0$ = permittivity of free space = 8.854  $\times {10}^{-12}$  $C^2{N}^{-1}$  $m^{-2}$

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

Potential of the inner surface is given by

V =  $\frac{q}{C}$ =  $\frac{2.5\times {10}^{-6}}{5.55\mathrm{}\times {10}^{-9}}$ = 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  $\pi {\epsilon }_0$ r

= 4  $\times \pi \times$ 8.854  $\times {10}^{-12}\times 12\mathrm{}\times {10}^{-2}$ 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.

 

Q.2.31 Answer carefully:

(a) Two large conducting spheres carrying charges  ${\mathit{Q}}_1$ and  ${\mathit{Q}}_2$ are brought close to each other. Is the magnitude of electrostatic force between them exactly given by  ${\mathit{Q}}_1{\mathit{Q}}_2$ /4  $\mathit{\pi }{\mathit{\epsilon }}_0{\mathit{r}}^2$ , where r is the distance between their centres?

(b) If Coulomb’s law involved 1/  ${\mathit{r}}^3$ dependence (instead of 1/  ${\mathit{r}}^2$ ), 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).

Ans.2.31 The force between two conducting spheres is not exactly given by the expression  ${\mathit{Q}}_1{\mathit{Q}}_2$ /4  $\mathit{\pi }{\mathit{\epsilon }}_0{\mathit{r}}^2$ , 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.

 

Q.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).

Ans.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  $\mu C=3.5\times {10}^{-6}$ C

Capacitance of a co-axial cylinder of radii  $r_1$ and  $r_2$ is given by the relation

C =  $\frac{2\pi {\epsilon }_{0}l}{{\log }_e\frac{r_1}{r_2}}$ , where  ${\epsilon }_0$ = permittivity of free space = 8.854  $\times {10}^{-12}$  $C^2{N}^{-1}$  $m^{-2}$

C =  $\frac{2\pi \times 8.854\mathrm{}\times {10}^{-12}\times 0.15}{{\log }_e\frac{0.015}{0.014}}$ = 1.21  $\times {10}^{-10}$ F

Potential difference of the inner cylinder is given by

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

 

Q.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?

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

Dielectric constant of a material,  ${ϵ}_r$ = 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 =  $\frac{{\epsilon }_0{ϵ}_{r}A}{d}$ , where  ${\epsilon }_0$ = permittivity of free space = 8.854  $\times {10}^{-12}$  $C^2{N}^{-1}$  $m^{-2}$

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

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

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

 

Q.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.

Ans.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.

 

Q.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.

Ans.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.

 

Q.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.)

Ans.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.