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Question

(C) Let the capacitance of each identical capacitor be $C$. Initially,capacitor $1$ has capacitance $C$ and capacitor $2$ has capacitance $KC$. Since

Vedclass · Accepted Answer

$\frac{Q'_2}{Q_2} = \frac{K + 1}{2K}$

Vedclass · Answer

(C) Let the capacitance of each identical capacitor be $C$. Initially,capacitor $1$ has capacitance $C$ and capacitor $2$ has capacitance $KC$. Since they are in series,the equivalent capacitance is $C_{eq} = \frac{C \cdot KC}{C + KC} = \frac{KC}{K+1}$. The charge on each capacitor is $Q_1 = Q_2 = Q_{eq} = C_{eq} V_0 = \frac{KCV_0}{K+1}$. After removing the dielectric,both capacitors have capacitance $C$. The new equivalent capacitance is $C'_{eq} = \frac{C \cdot C}{C + C} = \frac{C}{2}$. The new charge on each capacitor is $Q'_1 = Q'_2 = Q'_{eq} = C'_{eq} V_0 = \frac{CV_0}{2}$. Now,calculating the ratio: $\frac{Q'_1}{Q_1} = \frac{Q'_2}{Q_2} = \frac{CV_0/2}{KCV_0/(K+1)} = \frac{K+1}{2K}$. Comparing this with the given options,option $C$ is correct.

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