A parallel-plate capacitor with plate area A | vedclass

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Question

(A) The capacitor can be viewed as two parts in parallel. One part is an air-filled capacitor with area $A/2$ and separation $d$. Its capacitance is $

Vedclass · Accepted Answer

$\frac{\varepsilon_{0} A}{d} \left( \frac{1}{2} + \frac{K_{1} K_{2}}{K_{1} + K_{2}} \right)$

Vedclass · Answer

(A) The capacitor can be viewed as two parts in parallel. One part is an air-filled capacitor with area $A/2$ and separation $d$. Its capacitance is $C_{1} = \frac{\varepsilon_{0} (A/2)}{d} = \frac{\varepsilon_{0} A}{2d}$. The other part consists of two dielectric slabs of area $A/2$ and thickness $d/2$ placed in series. The capacitance of the slab with $K_{1}$ is $C_{2} = \frac{K_{1} \varepsilon_{0} (A/2)}{d/2} = \frac{K_{1} \varepsilon_{0} A}{d}$. The capacitance of the slab with $K_{2}$ is $C_{3} = \frac{K_{2} \varepsilon_{0} (A/2)}{d/2} = \frac{K_{2} \varepsilon_{0} A}{d}$. Since these two are in series,their equivalent capacitance $C_{s}$ is given by $\frac{1}{C_{s}} = \frac{1}{C_{2}} + \frac{1}{C_{3}} = \frac{d}{K_{1} \varepsilon_{0} A} + \frac{d}{K_{2} \varepsilon_{0} A} = \frac{d}{\varepsilon_{0} A} \left( \frac{1}{K_{1}} + \frac{1}{K_{2}} \right) = \frac{d}{\varepsilon_{0} A} \left( \frac{K_{1} + K_{2}}{K_{1} K_{2}} \right)$. Thus,$C_{s} = \frac{\varepsilon_{0} A}{d} \left( \frac{K_{1} K_{2}}{K_{1} + K_{2}} \right)$. The total capacitance $C_{eq}$ is the parallel combination of $C_{1}$ and $C_{s}$: $C_{eq} = C_{1} + C_{s} = \frac{\varepsilon_{0} A}{2d} + \frac{\varepsilon_{0} A}{d} \left( \frac{K_{1} K_{2}}{K_{1} + K_{2}} \right) = \frac{\varepsilon_{0} A}{d} \left( \frac{1}{2} + \frac{K_{1} K_{2}}{K_{1} + K_{2}} \right)$.

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