A parallel plate capacitor has the space between its plates filled by two slabs of thickness $ \frac{d}{2} $ each and dielectric constant $ K_{1} $ and $ K_{2} . d $ is the plate separation of the capacitor. The capacity of the capacitor is :

Question

A parallel plate capacitor has the space between its plates filled by two slabs of thickness $ \frac{d}{2} $ each and dielectric constant $ K_{1} $ and $ K_{2} . d $ is the plate separation of the capacitor. The capacity of the capacitor is :, $ \frac{2 \varepsilon_{0} d}{A}\left(\frac{K_{1}+K_{2}}{K_{1} K_{2}}\right) $, $ \frac{2 \varepsilon_{0} A}{d}\left(\frac{K_{1} K_{2}}{K_{1}+K_{2}}\right) $, $ \frac{2 \varepsilon_{0} A}{A}\left(K_{1}+K_{2}\right) $, $ \frac{2 \varepsilon_{0} A}{d}\left(\frac{K_{1}+K_{2}}{K_{1} K_{2}}\right) $

MARKS App · Accepted Answer

Let us understand this such that two capacitors are connected in series with K 1 and K 2 as dielectric constant of the dielectric mediums present in between the plates. See the figure down below for better understanding. C 1 = K 1 ε 0 A ( d 2 ) = 2 K 1 ε 0 A d C 2 = K 2 ε 0 A ( d 2 ) = 2 K 2 ε 0 A d Now since both C 1 and C 2 are arranged in series, we apply the formula for series combination of capacitor: 1 C s = 1 C 1 + 1 C 2 = d 2 K 1 ε 0 A + d 2 K 2 ε 0 A = d 2 ε 0 A K 1 + K 2 K 1 K 2 C s = 2 ε 0 A d K 1 K 2 K 1 + K 2

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