JEE MAIN - Physics (2024 - 8th April Evening Shift - No. 11)
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 $$\epsilon_0=8.834 \times 10^{-12} \mathrm{~F} / \mathrm{m}$$)
1.50
0.66
1.33
1
Explanation
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$$\begin{aligned} & \frac{\in_0 A}{d}=\frac{\epsilon_0 A}{\frac{d}{k}+\frac{0.2}{1}} \\ & \Rightarrow \quad 0.6-0.2=\frac{0.6}{k} \\ & \quad k=\frac{0.6}{0.4}=1.5 \end{aligned}$$
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