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JEE Mains · Physics · STD 12 - 2. Electric potential and capacitance

\(a\) બાજુવાળી ચોરસ પ્લેટમાથી એક કેપેસીટર બનાવેલ છે જે એક બીજા સાથે ખૂબ નાનો ખૂણો \(\alpha\) બનાવે છે. તો તેનો કેપેસીટન્સ કેટલો થાય?

  1. A \(\frac{\varepsilon_{0} a^{2}}{d}\left(1-\frac{3 \alpha a}{2 d}\right)\)
  2. B \(\frac{\varepsilon_{0} a^{2}}{d}\left(1-\frac{\alpha a}{4 d}\right)\)
  3. C \(\frac{\varepsilon_{0} {a}^{2}}{\mathrm{d}}\left(1+\frac{\alpha {a}}{\mathrm{d}}\right)\)
  4. D \(\frac{\varepsilon_{0} a^{2}}{d}\left(1-\frac{\alpha a}{2 d}\right)\)
Verified Solution

Answer & Solution

Correct Answer

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

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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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