In the given circuit,a charge of +80 mu C is | vedclass

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(C) Let the potential of the node between the $4 \ \mu F$ capacitor and the parallel combination be $V$. The $4 \ \mu F$ capacitor is in series with t

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$+48$

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(C) Let the potential of the node between the $4 \ \mu F$ capacitor and the parallel combination be $V$. The $4 \ \mu F$ capacitor is in series with the parallel combination of $2 \ \mu F$ and $3 \ \mu F$ capacitors.The charge on the $4 \ \mu F$ capacitor is $Q = 80 \ \mu C$. The potential difference across the $4 \ \mu F$ capacitor is $V_{4} = Q / C = 80 \ \mu C / 4 \ \mu F = 20 \ V$.Since the bottom plates of the $2 \ \mu F$ and $3 \ \mu F$ capacitors are grounded (potential $= 0 \ V$),the potential at the junction node is $V = -20 \ V$ relative to the top plate of the $4 \ \mu F$ capacitor,but using the node method where the charge $80 \ \mu C$ is supplied to the $4 \ \mu F$ capacitor,the potential at the junction $x$ relative to ground is determined by the charge distribution.Using the charge conservation at the junction node $x$:$Q_{total} = Q_{2\mu F} + Q_{3\mu F} = 80 \ \mu C$$x(2 \ \mu F) + x(3 \ \mu F) = 80 \ \mu C$$5x = 80 \ \mu C \implies x = 16 \ V$The charge on the $3 \ \mu F$ capacitor is $Q_{3} = C_{3} \cdot x = 3 \ \mu F \cdot 16 \ V = 48 \ \mu C$.

In the given circuit,a charge of $+80 \ \mu C$ is given to the upper plate of the $4 \ \mu F$ capacitor. Find the charge on the upper plate of the $3 \ \mu F$ capacitor in the steady state in $\mu C$.

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