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2. Electric Potential and Capacitance
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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
$1$
(JEE MAIN-2024)
Solution
$\frac{\mathrm{A} \varepsilon_{\mathrm{o}}}{\mathrm{d}}=\frac{\mathrm{A} \varepsilon_0}{\left(0.2+\frac{\mathrm{d}}{\mathrm{k}}\right)}$
$0.6=0.2+\frac{0.6}{\mathrm{k}}$
$\mathrm{k}=\frac{3}{2}$
Standard 12
Physics
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