Question bank/JEE Main/Physics

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

A$\frac{\varepsilon_{0}a^{2}}{d}(1 - \frac{\alpha a}{2d})$B$\frac{\varepsilon_{0}a^{2}}{d}(1 + \frac{\alpha a}{2d})$C$\frac{\varepsilon_{0}a^{2}}{d}(1 - \frac{\alpha a}{4d})$D$\frac{\varepsilon_{0}a^{2}}{d}(1 - \frac{3\alpha a}{2d})$

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$\Rightarrow c = \frac{\varepsilon_{0}a}{\alpha}\lbrack ln(d + \alpha x)\rbrack_{0}^{a}$
$= \frac{\varepsilon_{0}a}{\alpha}In(1 + \frac{\alpha a}{d}) \approx \frac{\varepsilon_{0}a^{2}}{d}(1 - \frac{\alpha a}{2d})$

Source: Grass-G / P-19 Physics paper 28 Nov Fourth.docx