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
JEE MAIN 2020, Diffcult
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Assume small element $dx$ at a distance $x$ from left end
Capacitance for small element $dx$ is
$\mathrm{d} \mathrm{C}=\frac{\varepsilon_{0} \mathrm{a} \mathrm{d} \mathrm{x}}{\mathrm{d}+\mathrm{x} \alpha}$
$C=\int_{0}^{a} \frac{\varepsilon_{0} a d x}{d+x \alpha}$
$=\left.\frac{\varepsilon_{0} a}{\alpha} \ln \left(\frac{1+\alpha a}{d}\right)\right|_{0} ^{a} \quad\left(\ln (1+x) \approx x-\frac{x^{2}}{2}\right)$
$=\frac{\varepsilon_{0} \mathrm{a}^{2}}{\mathrm{d}}\left(1-\frac{\alpha \mathrm{a}}{2 \mathrm{d}}\right)$
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