Capacitor A has a capacitance of 1 mu F. It i | vedclass

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(D) Initial capacitance of $A$ with dielectric: $C_A = K 	imes C_0 = 15 	imes 1\ \mu F = 15\ \mu F$.Charge on $A$: $Q_A = C_A 	imes V = 15\ \mu F \

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$800$

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(D) Initial capacitance of $A$ with dielectric: $C_A = K 	imes C_0 = 15 	imes 1\ \mu F = 15\ \mu F$.Charge on $A$: $Q_A = C_A 	imes V = 15\ \mu F 	imes 100\ V = 1500\ \mu C$.Charge on $B$: $Q_B = C_B 	imes V = 1\ \mu F 	imes 100\ V = 100\ \mu C$.Total charge $Q_{total} = Q_A + Q_B = 1500\ \mu C + 100\ \mu C = 1600\ \mu C$.After removing the dielectric,the capacitance of $A$ becomes $C'_A = 1\ \mu F$. The capacitance of $B$ remains $C'_B = 1\ \mu F$.When connected in parallel,the equivalent capacitance is $C_{eq} = C'_A + C'_B = 1\ \mu F + 1\ \mu F = 2\ \mu F$.The common potential $V_{cm}$ is given by $V_{cm} = \frac{Q_{total}}{C_{eq}} = \frac{1600\ \mu C}{2\ \mu F} = 800\ V$.

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