Capacitance

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

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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

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(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

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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…

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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.

Click to See Answer :
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…

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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

Click to See Answer :
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…

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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}} $

Click to See Answer :
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…

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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

Numerical

(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}} $

Click to See Answer :
Ans: (b)
Sol: Force acting between plates of capacitor is attractive & it is

$ \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} $