A parallel plate capacitor of area A,plate se | vedclass

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(B) The capacitor can be viewed as a combination of three capacitors. The top half consists of two capacitors in parallel with areas $A/2$ and separat

Vedclass · Accepted Answer

$\frac{1}{k} = \frac{1}{k_1 + k_2} + \frac{1}{2k_3}$

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(B) The capacitor can be viewed as a combination of three capacitors. The top half consists of two capacitors in parallel with areas $A/2$ and separation $d/2$,having capacitances $C_1 = \frac{k_1 \epsilon_0 (A/2)}{d/2} = \frac{k_1 \epsilon_0 A}{d}$ and $C_2 = \frac{k_2 \epsilon_0 (A/2)}{d/2} = \frac{k_2 \epsilon_0 A}{d}$.These two are in parallel,so their equivalent capacitance is $C_{12} = C_1 + C_2 = \frac{\epsilon_0 A}{d} (k_1 + k_2)$.The bottom half is a capacitor with area $A$ and separation $d/2$,having capacitance $C_3 = \frac{k_3 \epsilon_0 A}{d/2} = \frac{2k_3 \epsilon_0 A}{d}$.Since $C_{12}$ and $C_3$ are in series,the total capacitance $C$ is given by $\frac{1}{C} = \frac{1}{C_{12}} + \frac{1}{C_3}$.Substituting the values: $\frac{1}{C} = \frac{d}{\epsilon_0 A (k_1 + k_2)} + \frac{d}{2k_3 \epsilon_0 A} = \frac{d}{\epsilon_0 A} \left( \frac{1}{k_1 + k_2} + \frac{1}{2k_3} \right)$.For a single dielectric $k$,$C = \frac{k \epsilon_0 A}{d}$,so $\frac{1}{C} = \frac{d}{k \epsilon_0 A}$.Comparing the two expressions,we get $\frac{1}{k} = \frac{1}{k_1 + k_2} + \frac{1}{2k_3}$.

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