A capacitor,C_1 = 6 mu F is charged to a pot | vedclass

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(B) Initially,the charge on capacitor $C_1$ is $q_{initial} = C_1 V_0 = 6 \ \mu F 	imes 5 \ V = 30 \ \mu C$. The charge on $C_2$ is $0 \ \mu C$.When

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$q_1 = 10 \ \mu C, q_2 = 20 \ \mu C$

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(B) Initially,the charge on capacitor $C_1$ is $q_{initial} = C_1 V_0 = 6 \ \mu F 	imes 5 \ V = 30 \ \mu C$. The charge on $C_2$ is $0 \ \mu C$.When the switch $S$ is closed,the charge redistributes until both capacitors reach a common potential $V_c$. By the law of conservation of charge,the total charge remains constant:$q_{total} = q_1 + q_2 = 30 \ \mu C + 0 \ \mu C = 30 \ \mu C$.At equilibrium,$q_1 = C_1 V_c$ and $q_2 = C_2 V_c$. Since they are connected in parallel,$V_c = \frac{q_{total}}{C_1 + C_2} = \frac{30 \ \mu C}{6 \ \mu F + 12 \ \mu F} = \frac{30}{18} \ V = \frac{5}{3} \ V$.Now,calculate the final charges:$q_1 = C_1 V_c = 6 \ \mu F 	imes \frac{5}{3} \ V = 10 \ \mu C$.$q_2 = C_2 V_c = 12 \ \mu F 	imes \frac{5}{3} \ V = 20 \ \mu C$.

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