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Question

(D) The capacitor can be viewed as two capacitors in series,each with plate area $A$ and plate separation $d/2$.The capacitance of the first capacitor

Vedclass · Accepted Answer

$\frac{{2{\varepsilon _0}A}}{d}\left( {\frac{{{K_1}{K_2}}}{{{K_1} + {K_2}}}} \right)$

Vedclass · Answer

(D) The capacitor can be viewed as two capacitors in series,each with plate area $A$ and plate separation $d/2$.The capacitance of the first capacitor is $C_1 = \frac{K_1 \varepsilon_0 A}{d/2} = \frac{2 K_1 \varepsilon_0 A}{d}$.The capacitance of the second capacitor is $C_2 = \frac{K_2 \varepsilon_0 A}{d/2} = \frac{2 K_2 \varepsilon_0 A}{d}$.Since they are in series,the equivalent capacitance $C_{eq}$ is given by $\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2}$.$\frac{1}{C_{eq}} = \frac{d}{2 K_1 \varepsilon_0 A} + \frac{d}{2 K_2 \varepsilon_0 A} = \frac{d}{2 \varepsilon_0 A} \left( \frac{1}{K_1} + \frac{1}{K_2} \right) = \frac{d}{2 \varepsilon_0 A} \left( \frac{K_1 + K_2}{K_1 K_2} \right)$.Therefore,$C_{eq} = \frac{2 \varepsilon_0 A}{d} \left( \frac{K_1 K_2}{K_1 + K_2} \right)$.

Two dielectric slabs of dielectric constants $K_1$ and $K_2$ are filled between the plates of a parallel plate capacitor as shown in the figure. If the area of each plate is $A$ and the separation between the plates is $d$,what is the equivalent capacitance of the capacitor?

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