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

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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}] $

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