## 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 ………….

Click to See Solution :
Ans: (1.3)
Sol: $\displaystyle \frac{1}{C} = \frac{1}{C_1} + \frac{1}{C_2}$

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

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

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.

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…

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

$V’ = \frac{CV + 0}{C+C}$

$V’ = \frac{V}{2}$

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

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

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