Two parallel plates of area A are separated b | vedclass

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Question

(A) The capacitor is divided into two parts in series,each with a plate separation of $d/2$ and area $A$.For the first part,the dielectric constant is

Vedclass · Accepted Answer

$\frac{4{\varepsilon _0}A}{3d}$

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(A) The capacitor is divided into two parts in series,each with a plate separation of $d/2$ and area $A$.For the first part,the dielectric constant is $k_1 = 1$,so the capacitance is $C_1 = \frac{k_1 \varepsilon_0 A}{d/2} = \frac{1 \cdot \varepsilon_0 A}{d/2} = \frac{2 \varepsilon_0 A}{d}$.For the second part,the dielectric constant is $k_2 = 2$,so the capacitance is $C_2 = \frac{k_2 \varepsilon_0 A}{d/2} = \frac{2 \cdot \varepsilon_0 A}{d/2} = \frac{4 \varepsilon_0 A}{d}$.Since they are in series,the equivalent capacitance $C_{eq}$ is given by $\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2}$.Substituting the values: $\frac{1}{C_{eq}} = \frac{d}{2 \varepsilon_0 A} + \frac{d}{4 \varepsilon_0 A} = \frac{2d + d}{4 \varepsilon_0 A} = \frac{3d}{4 \varepsilon_0 A}$.Therefore,$C_{eq} = \frac{4 \varepsilon_0 A}{3d}$.

Two parallel plates of area $A$ are separated by two different dielectrics as shown in the figure. Find the resultant capacitance.

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