A parallel plate capacitor with square plates | vedclass

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(C) The capacitor can be viewed as two capacitors in series,each of thickness $d/2$. For the first half (thickness $d/2$),we have two capacitors in pa

Vedclass · Accepted Answer

$K = \frac{(K_1 + K_3)(K_2 + K_4)}{K_1 + K_2 + K_3 + K_4}$

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(C) The capacitor can be viewed as two capacitors in series,each of thickness $d/2$. For the first half (thickness $d/2$),we have two capacitors in parallel with areas $L^2/2$ each: $C_1 = \frac{\varepsilon_0 K_1 (L^2/2)}{d/2} + \frac{\varepsilon_0 K_3 (L^2/2)}{d/2} = \frac{\varepsilon_0 L^2}{d} (K_1 + K_3)$. For the second half (thickness $d/2$),we have two capacitors in parallel with areas $L^2/2$ each: $C_2 = \frac{\varepsilon_0 K_2 (L^2/2)}{d/2} + \frac{\varepsilon_0 K_4 (L^2/2)}{d/2} = \frac{\varepsilon_0 L^2}{d} (K_2 + K_4)$. Since these two halves are in series,the equivalent capacitance $C$ is given by: $\frac{1}{C} = \frac{1}{C_1} + \frac{1}{C_2} = \frac{d}{\varepsilon_0 L^2 (K_1 + K_3)} + \frac{d}{\varepsilon_0 L^2 (K_2 + K_4)}$. Also,$C = \frac{\varepsilon_0 K L^2}{d}$,so: $\frac{d}{\varepsilon_0 K L^2} = \frac{d}{\varepsilon_0 L^2} \left( \frac{1}{K_1 + K_3} + \frac{1}{K_2 + K_4} \right)$. $\frac{1}{K} = \frac{(K_2 + K_4) + (K_1 + K_3)}{(K_1 + K_3)(K_2 + K_4)}$. Therefore,$K = \frac{(K_1 + K_3)(K_2 + K_4)}{K_1 + K_2 + K_3 + K_4}$.

$A$ parallel plate capacitor with square plates of side $L$ is filled with four dielectrics of dielectric constants $K_1, K_2, K_3, K_4$ arranged as shown in the figure. The effective dielectric constant $K$ will be

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