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(D) The capacitors $C_1 = 2\,\mu F$ and $C_2 = 3\,\mu F$ are connected in series between points $A$ and $B$.Given $V_A = 1000\,V$ and $V_B = 0\,V$,the

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$400$

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(D) The capacitors $C_1 = 2\,\mu F$ and $C_2 = 3\,\mu F$ are connected in series between points $A$ and $B$.Given $V_A = 1000\,V$ and $V_B = 0\,V$,the potential difference across the combination is $V_{AB} = V_A - V_B = 1000\,V$.The equivalent capacitance $C_{eq}$ is given by $\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} = \frac{1}{2} + \frac{1}{3} = \frac{5}{6}$,so $C_{eq} = \frac{6}{5}\,\mu F$.The charge $Q$ on each capacitor in series is $Q = C_{eq} 	imes V_{AB} = \frac{6}{5} 	imes 1000 = 1200\,\mu C$.The potential difference across the $2\,\mu F$ capacitor is $V_A - V_C = \frac{Q}{C_1} = \frac{1200\,\mu C}{2\,\mu F} = 600\,V$.Substituting $V_A = 1000\,V$,we get $1000 - V_C = 600\,V$.Therefore,$V_C = 1000 - 600 = 400\,V$.

In the given circuit,what is the electric potential at point $C$ in $volts$?

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