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…

Question

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 $\epsilon_0=8.834 	imes 10^{-12} \mathrm{~F} / \mathrm{m}$ )

Aicharya · Accepted Answer

$\begin{aligned} & \frac{\mathrm{A} \varepsilon_{\mathrm{o}}}{\mathrm{d}}=\frac{\mathrm{A} \varepsilon_{\mathrm{o}}}{\left(0.2+\frac{\mathrm{d}}{\mathrm{k}}ight)} \ & 0.6=0.2+\frac{0.6}{\mathrm{k}} \ & \mathrm{k}=\frac{3}{2}\end{aligned}$

A capacitor has air as dielectric medium and two conducting plates of area $12 \mathrm{~cm}^2$ and they are…

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