JEE Mains · Physics · STD 12 - Electric potential and capacitance
- JEE Mains 2020
- Hard
- MCQ
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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Answer & Solution
Correct Answer
(D) \(\frac{\varepsilon_{0} a^{2}}{d}\left(1-\frac{\alpha a}{2 d}\right)\)
Step-by-step Solution
Detailed explanation
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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