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(B) $1$. Initially,capacitor $B$ is charged to $V_0 = 30\,V$. The charge on $B$ is $Q_0 = C_0 V_0 = 30\,C_0$.$2$. When capacitor $B$ is connected to c

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$A$ are $25\,C_0$ and $5\,V$

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(B) $1$. Initially,capacitor $B$ is charged to $V_0 = 30\,V$. The charge on $B$ is $Q_0 = C_0 V_0 = 30\,C_0$.$2$. When capacitor $B$ is connected to capacitor $A$,charge flows until they reach a common potential $V$. Let the final charge on $A$ be $Q_A$ and on $B$ be $Q_B$.$3$. The capacitance of $A$ is $C_A = K C_0 = V C_0$. The charge on $A$ is $Q_A = C_A V = (V C_0) V = C_0 V^2$.$4$. The charge on $B$ is $Q_B = C_0 V$.$5$. By conservation of charge,$Q_A + Q_B = Q_0$,so $C_0 V^2 + C_0 V = 30\,C_0$.$6$. Simplifying,$V^2 + V - 30 = 0$,which factors to $(V + 6)(V - 5) = 0$. Since $V > 0$,$V = 5\,V$.$7$. The charge on $A$ is $Q_A = C_0 V^2 = C_0 (5)^2 = 25\,C_0$.

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