JEE Mains · Physics · STD 12 - 2. Electric potential and capacitance
A capacitor has air as dielectric medium and two conducting plates of area \(12 \mathrm{~cm}^2\) and they are \(0.6 \mathrm{~cm}\) apart. When a slab of dielectric having area \(12 \mathrm{~cm}^2\) and \(0.6 \mathrm{~cm}\) thickness is inserted between the plates, one of the conducting plates has to be moved by \(0.2 \mathrm{~cm}\) to keep the capacitance same as in previous case. The dielectric constant of the slab is _______. (Given \(\left.\epsilon_0=8.834 \times 10^{-12} \mathrm{~F} / \mathrm{m}\right)\)
- A \(1.50\)
- B \(1.33\)
- C \(0.66\)
- D \(1\)
Answer & Solution
Correct Answer
(A) \(1.50\)
Step-by-step Solution
Detailed explanation
\(\frac{\mathrm{A} \varepsilon_{\mathrm{o}}}{\mathrm{d}}=\frac{\mathrm{A} \varepsilon_0}{\left(0.2+\frac{\mathrm{d}}{\mathrm{k}}\right)}\) \(0.6=0.2+\frac{0.6}{\mathrm{k}}\) \(\mathrm{k}=\frac{3}{2}\)
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