A capacitor of capacitance C_1 = 1 mu F is c | vedclass

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(A) Initially,when $C_1$ is connected to a $9 \ V$ battery,the charge on it is $q_0 = C_1 V = 1 \ \mu F 	imes 9 \ V = 9 \ \mu C$.When $C_1$ is connec

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$4.5 	imes 10^{-6} \ C$

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(A) Initially,when $C_1$ is connected to a $9 \ V$ battery,the charge on it is $q_0 = C_1 V = 1 \ \mu F 	imes 9 \ V = 9 \ \mu C$.When $C_1$ is connected to $C_2$ and $C_3$ in parallel,the total charge $q_0$ is redistributed among the three capacitors such that they all have the same potential difference $V'$.By the law of conservation of charge,$q_1 + q_2 + q_3 = q_0 = 9 \ \mu C$.Since they are in parallel,$V' = \frac{q_1}{C_1} = \frac{q_2}{C_2} = \frac{q_3}{C_3}$.Substituting the values,$\frac{q_1}{1} = \frac{q_2}{2} = \frac{q_3}{3} = V'$.Thus,$q_1 = V'$,$q_2 = 2V'$,and $q_3 = 3V'$.Substituting these into the conservation equation: $V' + 2V' + 3V' = 9 \ \mu C$.$6V' = 9 \ \mu C \Rightarrow V' = 1.5 \ V$.The charge on $C_3$ is $q_3 = C_3 V' = 3 \ \mu F 	imes 1.5 \ V = 4.5 \ \mu C = 4.5 	imes 10^{-6} \ C$.

$A$ capacitor of capacitance $C_1 = 1 \ \mu F$ is charged using a $9 \ V$ battery. $C_1$ is then removed from the battery and connected to capacitors $C_2$ and $C_3$ of $2 \ \mu F$ and $3 \ \mu F$ respectively,as shown in the figure. Find the charge on $C_3$ after equilibrium is reached.

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