A composite parallel plate capacitor is made | vedclass

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Question

(D) The composite capacitor acts as two capacitors $C_{1}$ and $C_{2}$ in series.$C_{1} = \frac{K_{1} \epsilon_{0} A}{t_{1}} = \frac{3 \epsilon_{0} A}

Vedclass · Accepted Answer

$60$

Vedclass · Answer

(D) The composite capacitor acts as two capacitors $C_{1}$ and $C_{2}$ in series.$C_{1} = \frac{K_{1} \epsilon_{0} A}{t_{1}} = \frac{3 \epsilon_{0} A}{0.5 	imes 10^{-3}} = 6000 \epsilon_{0} A$$C_{2} = \frac{K_{2} \epsilon_{0} A}{t_{2}} = \frac{4 \epsilon_{0} A}{1 	imes 10^{-3}} = 4000 \epsilon_{0} A$Since they are in series,the charge $q$ on each capacitor is the same.The potential drop across each capacitor is inversely proportional to its capacitance: $V_{1} + V_{2} = 100 	ext{ V}$.Also,$\frac{V_{1}}{V_{2}} = \frac{C_{2}}{C_{1}} = \frac{4000 \epsilon_{0} A}{6000 \epsilon_{0} A} = \frac{2}{3}$.$V_{1} = \frac{2}{5} 	imes 100 = 40 	ext{ V}$ and $V_{2} = \frac{3}{5} 	imes 100 = 60 	ext{ V}$.Taking the bottom plate at $0 	ext{ V}$,the potential of the foil $F$ is the potential drop across $C_{2}$,which is $60 	ext{ V}$.

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