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(D) Consider a small strip of width $dx$ at a distance $x$ from the left end. The separation between the plates at this distance $x$ is $d' = d + x\al

Vedclass · Accepted Answer

$\frac{\varepsilon_{0} a^{2}}{d}\left(1-\frac{\alpha a}{2 d}\right)$

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(D) Consider a small strip of width $dx$ at a distance $x$ from the left end. The separation between the plates at this distance $x$ is $d' = d + x\alpha$.The capacitance of this small strip is $dC = \frac{\varepsilon_0 a dx}{d + x\alpha}$.To find the total capacitance $C$,we integrate from $x = 0$ to $x = a$:$C = \int_{0}^{a} \frac{\varepsilon_0 a dx}{d + x\alpha} = \frac{\varepsilon_0 a}{\alpha} [\ln(d + x\alpha)]_{0}^{a} = \frac{\varepsilon_0 a}{\alpha} \ln\left(\frac{d + a\alpha}{d}\right) = \frac{\varepsilon_0 a}{\alpha} \ln\left(1 + \frac{a\alpha}{d}\right)$.Using the Taylor series expansion $\ln(1 + y) \approx y - \frac{y^2}{2}$ for small $y = \frac{a\alpha}{d}$:$C \approx \frac{\varepsilon_0 a}{\alpha} \left(\frac{a\alpha}{d} - \frac{1}{2} \left(\frac{a\alpha}{d}\right)^2\right) = \frac{\varepsilon_0 a^2}{d} \left(1 - \frac{a\alpha}{2d}\right)$.

$A$ capacitor is made of two square plates each of side $a$ making a very small angle $\alpha$ between them,as shown in figure. The capacitance will be close to

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