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

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(C) The $2\,\mu F$ and $3\,\mu F$ capacitors are connected in parallel. Their equivalent capacitance is $C_p = 2\,\mu F + 3\,\mu F = 5\,\mu F$.This pa

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$+ 48$

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(C) The $2\,\mu F$ and $3\,\mu F$ capacitors are connected in parallel. Their equivalent capacitance is $C_p = 2\,\mu F + 3\,\mu F = 5\,\mu F$.This parallel combination is in series with the $4\,\mu F$ capacitor.The total charge $Q = +80\,\mu C$ resides on the $4\,\mu F$ capacitor and is distributed across the parallel combination.Since the capacitors are in series, the charge on the equivalent $5\,\mu F$ capacitor is also $Q = +80\,\mu C$.The potential difference across the parallel combination is $V = \frac{Q}{C_p} = \frac{80\,\mu C}{5\,\mu F} = 16\, V$.Since the $3\,\mu F$ capacitor is in parallel, the potential difference across it is also $16\, V$.Therefore, the charge on the $3\,\mu F$ capacitor is $q = C 	imes V = 3\,\mu F 	imes 16\, V = 48\,\mu C$.

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

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