A parallel plate capacitor of area A and plat | vedclass

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Question

(D) The arrangement can be divided into three capacitors:$1$. Capacitor $C_1$ (air) with area $A/2$ and separation $d/2$: $C_1 = \frac{\varepsilon_0 (

Vedclass · Accepted Answer

$\frac{K(K+3) \varepsilon_0 A}{2(K+1) d}$

Vedclass · Answer

(D) The arrangement can be divided into three capacitors:$1$. Capacitor $C_1$ (air) with area $A/2$ and separation $d/2$: $C_1 = \frac{\varepsilon_0 (A/2)}{d/2} = \frac{\varepsilon_0 A}{d}$.$2$. Capacitor $C_2$ (dielectric $K$) with area $A/2$ and separation $d/2$: $C_2 = \frac{K \varepsilon_0 (A/2)}{d/2} = \frac{K \varepsilon_0 A}{d}$.$3$. Capacitor $C_3$ (dielectric $K$) with area $A/2$ and separation $d$: $C_3 = \frac{K \varepsilon_0 (A/2)}{d} = \frac{K \varepsilon_0 A}{2d}$.$C_1$ and $C_2$ are in series,so their equivalent capacitance $C_{12} = \frac{C_1 C_2}{C_1 + C_2} = \frac{(\frac{\varepsilon_0 A}{d})(\frac{K \varepsilon_0 A}{d})}{\frac{\varepsilon_0 A}{d} + \frac{K \varepsilon_0 A}{d}} = \frac{K \varepsilon_0 A}{d(K+1)}$.$C_{12}$ is in parallel with $C_3$,so $C_{eq} = C_{12} + C_3 = \frac{K \varepsilon_0 A}{d(K+1)} + \frac{K \varepsilon_0 A}{2d} = \frac{K \varepsilon_0 A}{d} [\frac{1}{K+1} + \frac{1}{2}] = \frac{K \varepsilon_0 A}{d} [\frac{2 + K + 1}{2(K+1)}] = \frac{K(K+3) \varepsilon_0 A}{2(K+1)d}$.Thus,the correct option is $D$.

$A$ parallel plate capacitor of area $A$ and plate separation $d$ is filled with two dielectrics as shown. What is the equivalent capacitance of the arrangement?

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