JEE Mains · Physics · STD 12 - 2. Electric potential and capacitance
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
- A \(\frac{\varepsilon_{0} a^{2}}{d}\left(1-\frac{3 \alpha a}{2 d}\right)\)
- B \(\frac{\varepsilon_{0} a^{2}}{d}\left(1-\frac{\alpha a}{4 d}\right)\)
- C \(\frac{\varepsilon_{0} {a}^{2}}{\mathrm{d}}\left(1+\frac{\alpha {a}}{\mathrm{d}}\right)\)
- D \(\frac{\varepsilon_{0} a^{2}}{d}\left(1-\frac{\alpha a}{2 d}\right)\)
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 \(\mathrm{d} \mathrm{C}=\frac{\varepsilon_{0} \mathrm{a} \mathrm{d} \mathrm{x}}{\mathrm{d}+\mathrm{x} \alpha}\) \(C=\int_{0}^{a} \frac{\varepsilon_{0} a d x}{d+x \alpha}\)…
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