In the figure,a capacitor is filled with diel | vedclass

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(D) The capacitor can be divided into three parts. Let the total area be $A$ and distance be $d$.$1$. The left side consists of two capacitors $C_1$ a

Vedclass · Accepted Answer

$\frac{{A{\varepsilon _0}}}{d}\left( {\frac{{{k_1}{k_2}}}{{{k_1} + {k_2}}} + \frac{{{k_3}}}{2}} \right)$

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(D) The capacitor can be divided into three parts. Let the total area be $A$ and distance be $d$.$1$. The left side consists of two capacitors $C_1$ and $C_2$ in series,each with area $A/2$ and distance $d/2$.$C_1 = \frac{K_1 \varepsilon_0 (A/2)}{d/2} = \frac{K_1 \varepsilon_0 A}{d}$$C_2 = \frac{K_2 \varepsilon_0 (A/2)}{d/2} = \frac{K_2 \varepsilon_0 A}{d}$Since they are in series,the equivalent capacitance $C_{12}$ is given by $\frac{1}{C_{12}} = \frac{1}{C_1} + \frac{1}{C_2} = \frac{d}{\varepsilon_0 A} (\frac{1}{K_1} + \frac{1}{K_2}) = \frac{d}{\varepsilon_0 A} (\frac{K_1 + K_2}{K_1 K_2})$.So,$C_{12} = \frac{\varepsilon_0 A}{d} (\frac{K_1 K_2}{K_1 + K_2})$.$2$. The right side is a capacitor $C_3$ with area $A/2$ and distance $d$.$C_3 = \frac{K_3 \varepsilon_0 (A/2)}{d} = \frac{K_3 \varepsilon_0 A}{2d}$.$3$. $C_{12}$ and $C_3$ are in parallel. The resultant capacitance $C_{eq} = C_{12} + C_3 = \frac{\varepsilon_0 A}{d} (\frac{K_1 K_2}{K_1 + K_2} + \frac{K_3}{2})$.

In the figure,a capacitor is filled with dielectrics of constants $K_1$,$K_2$,and $K_3$. The resultant capacitance is

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