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

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(B) The $4 \mu F$ capacitor is in series with the parallel combination of the $2 \mu F$ and $3 \mu F$ capacitors. When a charge of $+80 \mu C$ is plac

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$48$

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(B) The $4 \mu F$ capacitor is in series with the parallel combination of the $2 \mu F$ and $3 \mu F$ capacitors. When a charge of $+80 \mu C$ is placed on the upper plate of the $4 \mu F$ capacitor,an equal and opposite charge of $-80 \mu C$ is induced on its lower plate. This charge of $+80 \mu C$ is then distributed between the upper plates of the $2 \mu F$ and $3 \mu F$ capacitors,which are connected in parallel. Since the capacitors are in parallel,the charge $q$ distributes in the ratio of their capacitances: $q_1 = \left( \frac{C_1}{C_1 + C_2} ight) Q_{total}$ For the $3 \mu F$ capacitor: $q = \left( \frac{3 \mu F}{3 \mu F + 2 \mu F} ight) 	imes 80 \mu C$ $q = \left( \frac{3}{5} ight) 	imes 80 \mu C = 3 	imes 16 \mu C = 48 \mu C$.

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

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