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(C) The capacitor is divided into two parts in series,each with a plate separation of $d/2$ and area $A$.Let $C_1$ and $C_2$ be the capacitances of th

Vedclass · Accepted Answer

$\frac{2 \varepsilon_0 A}{d}\left(\frac{K_1 K_2}{K_1+K_2}\right)$

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(C) The capacitor is divided into two parts in series,each with a plate separation of $d/2$ and area $A$.Let $C_1$ and $C_2$ be the capacitances of the two parts:$C_1 = \frac{K_1 \varepsilon_0 A}{d/2} = \frac{2 K_1 \varepsilon_0 A}{d}$$C_2 = \frac{K_2 \varepsilon_0 A}{d/2} = \frac{2 K_2 \varepsilon_0 A}{d}$Since they are in series,the equivalent capacitance $C_{eq}$ is given by:$\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} = \frac{d}{2 K_1 \varepsilon_0 A} + \frac{d}{2 K_2 \varepsilon_0 A}$$\frac{1}{C_{eq}} = \frac{d}{2 \varepsilon_0 A} \left( \frac{1}{K_1} + \frac{1}{K_2} \right) = \frac{d}{2 \varepsilon_0 A} \left( \frac{K_1 + K_2}{K_1 K_2} \right)$$C_{eq} = \frac{2 \varepsilon_0 A}{d} \left( \frac{K_1 K_2}{K_1 + K_2} \right)$

$A$ parallel plate capacitor with plate area $A$ and separation $d$ is filled with two dielectric materials as shown in the figure. The dielectric constants are $K_1$ and $K_2$ respectively. The capacitance will be:

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