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

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(C) The total charge $Q = +80 \ \mu C$ is supplied to the $4 \ \mu F$ capacitor. This charge then distributes between the parallel combination of the

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$+48 \ \mu C$

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(C) The total charge $Q = +80 \ \mu C$ is supplied to the $4 \ \mu F$ capacitor. This charge then distributes between the parallel combination of the $2 \ \mu F$ and $3 \ \mu F$ capacitors.The equivalent capacitance of the parallel combination is $C_p = 2 \ \mu F + 3 \ \mu F = 5 \ \mu F$.The charge $Q$ is distributed across the parallel capacitors in proportion to their capacitances. The charge $q_3$ on the $3 \ \mu F$ capacitor is given by the charge division rule:$q_3 = \left( \frac{C_3}{C_2 + C_3} ight) \cdot Q$Substituting the given values:$q_3 = \left( \frac{3 \ \mu F}{2 \ \mu F + 3 \ \mu F} ight) 	imes 80 \ \mu C$$q_3 = \left( \frac{3}{5} ight) 	imes 80 \ \mu C$$q_3 = 3 	imes 16 \ \mu C = 48 \ \mu C$Thus,the charge on the upper plate of the $3 \ \mu F$ capacitor is $+48 \ \mu C$.

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

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