Physics · Capacitors and R-C Circuits

JEE Main 2024 — 8 April, Shift 2 — Question 40

A capacitor has air as dielectric medium and two conducting plates of area $12 \mathrm{~cm}^{2}$ and they are 0.6 cm apart. When a slab

of dielectric having area $12 \mathrm{~cm}^{2}$ and 0.6 cm thickness is inserted between the plates, one of the conducting plates has to

be moved by 0.2 cm to keep the capacitance same as in previous case. The dielectric constant of the slab is :

$\left(\right.$ Given $\left.\epsilon_{0}=8.834 \times 10^{-12} \mathrm{~F} / \mathrm{m}\right)$

Answer: A

$\frac{A \varepsilon_{o}}{d}=\frac{A \varepsilon_{o}}{\left(0.2+\frac{d}{k}\right)}$

$0.6=0.2+\frac{0.6}{\mathrm{k}}$

$\mathrm{k}=\frac{3}{2}$

Answer key and solution verified before publishing.

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