Capacitance

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} $

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

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 = V_ball – V_∞ =  V

V₂ = V_ball– 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 $\displaystyle C = 2 \pi \epsilon_0 K a$

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})} $

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

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

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

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

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 %

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

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.

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

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}} $

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}} $