Capacitance

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 %

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}] $

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

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}] $

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

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 $

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