A parallel plate capacitor with plate area A has separation d between… | JEE Main 2021 (26 Aug, Shift 2)

Q82JEE Main 2021 · 26 AUG · SHIFT 2Medium

Dielectrics

A parallel plate capacitor with plate area A has separation d between the plates. Two dielectric slabs of dielectric constant K1 and K2 of same area A/2 and thickness d/2 are inserted in the space between the plates. The capacitance of the capacitor will be given by:

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Answer: (a)

Left half (area $\frac{A}{2}$ , air, gap $d$ ): $\frac{\epsilon_0A}{2d}$ .
Right half: $K_1,K_2$ layers of $\frac{d}{2}$ in series: $\frac{\epsilon_0A/2}{\frac{d}{2K_1}+\frac{d}{2K_2}}=\frac{\epsilon_0A}{d}\cdot\frac{K_1K_2}{K_1+K_2}$
The two halves are in parallel: $C=\frac{\epsilon_0A}{d}\left(\frac{1}{2}+\frac{K_1K_2}{K_1+K_2}\right)$
So the answer is (a).

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