## Capacitor has square plates each of side ‘ l ‘ making an angle ‘ α ‘ with each other as shown. Then for small value of α ….

Q: Capacitor has square plates each of side ‘ l ‘ making an angle ‘ α ‘ with each other as shown. Then for small value of α , the capacitance ‘C’ is given by Solution : At one side, distance between plates d,

At another side,

distance = d + l sinα  ≅  d + l α

Mean distance between the plates

$\displaystyle = \frac{d + d + l \alpha}{2}$

$\displaystyle = d + \frac{l \alpha}{2}$

Capacity , $\displaystyle C = \frac{\epsilon_0 A}{d}$

$\displaystyle C = \frac{\epsilon_0 l^2 }{d + \frac{l \alpha}{2}}$

$\displaystyle C = \frac{\epsilon_0 l^2}{d} [1 + \frac{l \alpha}{2 d}]^{-1}$

$\displaystyle C = \frac{\epsilon_0 l^2}{d} [1 – \frac{l \alpha}{2 d}]$

## Find the capacitance of a system of two identical metal balls of radius a if the distance between their centers is equal to b…

Q: Find the capacitance of a system of two identical metal balls of radius a if the distance between their centres is equal to b, with b >> a. The system is located in a uniform dielectric with permittivity K.

Solution : Let q and –q be the charges on two balls. Then

V1 = Vball – V =  V

V2 = Vball– V = -V The potential difference between the balls

$\displaystyle = V_1 – V_2 = 2 V$

$\displaystyle = 2 \int_{a}^{b-a} E dr$

$\displaystyle = 2 \int_{a}^{b-a} (\frac{1}{4 \pi \epsilon_0 K }\frac{q}{r^2} ) dr$

$\displaystyle = \frac{2 q}{4 \pi \epsilon_0 K } \int_{a}^{b-a} \frac{1}{r^2} ) dr$

$\displaystyle = \frac{2 q}{4 \pi \epsilon_0 K } [\frac{1}{a} – \frac{1}{b-a}]$

$\displaystyle C = \frac{q}{V_1 – V_2}$

$\displaystyle C = \frac{q}{\frac{2 q}{4 \pi \epsilon_0 K } [\frac{1}{a} – \frac{1}{b-a}]}$

$\displaystyle C = \frac{2 \pi \epsilon_0 K a (b-a)}{(b-2a)}$

For b >> a, we can write $\displaystyle C = 2 \pi \epsilon_0 K a$

## Two conductors carrying equal and opposite charge produce a non uniform electric field along X – axis given by …..

Q: Two conductors carrying equal and opposite charge produce a non uniform electric field along X – axis given by $E = \frac{Q}{\epsilon_0 A}(1 + B x^2)$ ; where A and B are constants. Separation between the conductors along X-axis is x . Find the capacitance of the capacitor formed.

Solution : Potential difference between the conductors is given by $V = V_{+} – V_{-} = \int_{0}^{x} E dx$

$V = \int_{0}^{x} \frac{Q}{\epsilon_0 A}(1 + B x^2) dx$

$V = \frac{Q}{\epsilon_0 A}[ x + B \frac{x^3}{3}]_{0}^{x}$

Capacitance $C = \frac{Q}{V} = \frac{\epsilon_0 A}{x(1 + B \frac{x^2}{3})}$

## A metal slab of thickness, equal to half the distance between the plates is introduced between the plates of a parallel plate

Q: A metal slab of thickness, equal to half the distance between the plates is introduced between the plates of a parallel plate capacitor as shown. Find its capacity. Solution : When capacitor is partially filled with metal Slab $\displaystyle C = \frac{\epsilon_0 A}{d – t}$

$\displaystyle C = \frac{\epsilon_0 A}{d – \frac{d}{2}}$

$\displaystyle C = \frac{\epsilon_0 A}{d/2}$

$\displaystyle C = 2 \frac{\epsilon_0 A}{d}$

## Two capacitors with capacitance values C1 = 2000 ±10 pF and C2 = 3000 ±15 pF are connected in series ….

Q: Two capacitors with capacitance values C1 = 2000 ± 10 pF and C2 = 3000 ± 15 pF are connected in series . The voltage applied across this combination is C = 5.00 ±0.02 V . The percentage error in the calculation of the energy stored in this combination is ………….

Ans: (1.3)

Solution: $\displaystyle \frac{1}{C} = \frac{1}{C_1} + \frac{1}{C_2}$

$\displaystyle C = \frac{C_1 C_2}{C_1 + C_2}$

$\displaystyle -\frac{dC}{C^2} = -\frac{dC_1}{C_1^2} – \frac{dC_2}{C_2^2}$

dC = 6 pF

$\displaystyle U = \frac{1}{2}CV^2$

$\displaystyle \frac{dU}{U} \times 100 = (\frac{dC}{C} + \frac{2dV}{V})\times 100$

$\displaystyle \frac{dU}{U} \times 100 = (\frac{6}{1200} + 2\times \frac{0.02}{5})\times 100$

= 13 %

## A capacitor is made of two square plates each of side ‘ a ‘ making a very small angle α between them ….

Q: A capacitor is made of two square plates each of side ‘ a ‘ making a very small angle α between them , as shown in figure .The capacitance will be close to (a) $\displaystyle \frac{\epsilon_0 a^2}{d}(1-\frac{\alpha a}{4d})$

(b) $\displaystyle \frac{\epsilon_0 a^2}{d}(1 +\frac{\alpha a}{4d})$

(c) $\displaystyle \frac{\epsilon_0 a^2}{d}(1-\frac{\alpha a}{2d})$

(d) $\displaystyle \frac{\epsilon_0 a^2}{d}(1-\frac{3\alpha a}{2d})$

Click to See Solution :
Ans: (c)

Sol: Let us consider a small element of thickness dx at a distance x from left Capacitance of small element $\displaystyle dC = \frac{\epsilon_0 a dx}{d + x \alpha}$

$\displaystyle C = \int_{0}^{a} \frac{\epsilon_0 a dx}{d + x \alpha}$

$\displaystyle C = \frac{\epsilon_0 a}{\alpha}[ln(1+\frac{x \alpha}{d})]_{0}^{a}$

$\displaystyle C = \frac{\epsilon_0 a}{\alpha}[ln(1+\frac{a \alpha}{d}) – ln1]$

$\displaystyle C = \frac{\epsilon_0 a}{\alpha}ln(1+\frac{a \alpha}{d})$

Applying $ln(1+y) = y – \frac{y^2}{2} + \frac{y^3}{3} – …..$

$\displaystyle C = \frac{\epsilon_0 a}{\alpha}[ \frac{a \alpha}{d} – \frac{a^2 \alpha^2}{2d^2}]$

$\displaystyle C = \frac{\epsilon_0 a}{\alpha}\times \frac{a \alpha}{d}[ 1 – \frac{a \alpha}{2d}]$

$\displaystyle C = \frac{\epsilon_0 a^2}{d}(1-\frac{\alpha a}{2d})$

## Two conductors carrying equal and opposite charge produce a non uniform electric field along X – axis given by…

Q: Two conductors carrying equal and opposite charge produce a non uniform electric field along X – axis given by $E = \frac{Q}{\epsilon_0 A} (1 + B x^2)$  where A and B are constants. Separation between the conductors along X=axis is X. Find the capacitance of the capacitor formed.

Sol. Potential difference between the conductors is given by

$\large V = V_+ – V_- = \int_{0}^{x} E dx$

$\large V = \int_{0}^{x} \frac{Q}{\epsilon_0 A} (1 + B x^2) dx$

$\large V = \frac{Q}{\epsilon_0 A} [x + \frac{B x^3}{3}]_{0}^{x}$

$\large V = \frac{Q}{\epsilon_0 A} [x + \frac{B x^3}{3}]$

$\large C = \frac{Q}{V} = \frac{\epsilon_0 A}{x + \frac{B x^3}{3}}$

## A capacitor is charged by a battery . The battery is removed and another identical uncharged capacitor…

Q: A capacitor is charged by a battery . The battery is removed and another identical uncharged capacitor is connected in parallel . The total electrostatic energy of resulting system :

(a) Decreases by a factor 2

(b) Remains same

(c) Increases by a factor 2

(d) Increases by a factor 4

Ans: (a)
Sol: $\large U_i = \frac{1}{2}CV^2$

Common Potential $V’ = \frac{C_1V_1 + C_2 V_2}{C_1+C_2}$

$V’ = \frac{CV + 0}{C+C}$

$V’ = \frac{V}{2}$

$\large U_f = \frac{1}{2}(C+C)(\frac{V}{2})^2$

$\large U_f = \frac{1}{4}CV^2$

$\large \frac{U_f}{U_i} = \frac{1}{2}$

## A fully charged capacitor has a capacitance C. It is discharged through a small coil of resistance…

Q: A fully charged capacitor has a capacitance C. It is discharged through a small coil of resistance wire embedded in a thermally insulated block of specific heat capacity S and mass m. If the temperature of the block is raised by  ΔT, the potential difference V across the capacitor is

(a) $\displaystyle \sqrt{\frac{2 m C \Delta T}{S} }$

(b) $\displaystyle \frac{ m C \Delta T}{S}$

(c) $\displaystyle \frac{ m S \Delta T}{C}$

(d) $\displaystyle \sqrt{\frac{2 m S \Delta T}{C}}$

Ans: (d)
Sol: $\displaystyle \frac{1}{2}C V^2 = m S \Delta T$

$\displaystyle V = \sqrt{\frac{2 m S \Delta T}{C}}$

## In the given figure the capacitor of plate area A is charged upto charge q. The ratio of elongations…

Q: In the given figure the capacitor of plate area A is charged upto charge q. The ratio of elongations (neglect force of gravity) in springs C and D at equilibrium position is (a) $\displaystyle \frac{k_1}{k_2}$

(b) $\displaystyle \frac{k_2}{k_1}$

(c) $\displaystyle k_1 k_2$

(d) $\displaystyle \sqrt{ \frac{k_1}{k_2}}$

$\displaystyle F = \frac{q^2}{2 \epsilon_0 A}$
$\displaystyle k_1 x_1 = k_2 x_2$
$\displaystyle \frac{x_1}{x_2} = \frac{k_2}{k_1}$