In the given figure,each plate of the capacit | vedclass

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(C) In the steady state,the capacitor acts as an open circuit,so no current flows through the branch containing the capacitor (Line $(2)$).The circuit

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$\frac{CE R_2}{R_2 + r}$

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(C) In the steady state,the capacitor acts as an open circuit,so no current flows through the branch containing the capacitor (Line $(2)$).The circuit simplifies to a series combination of the battery $E$,its internal resistance $r$,and the resistor $R_2$.The current $i$ flowing through the circuit is given by $i = \frac{E}{R_2 + r}$.The potential difference $V$ across the resistor $R_2$ is $V = i R_2 = \frac{E R_2}{R_2 + r}$.Since the capacitor is connected in parallel with the resistor $R_2$,the potential difference across the capacitor is the same as the potential difference across $R_2$.Therefore,the charge $Q$ on the capacitor is $Q = C V = C \left( \frac{E R_2}{R_2 + r} \right) = \frac{C E R_2}{R_2 + r}$.

In the given figure,each plate of the capacitor with capacitance $C$ has a charge. Find the charge on the capacitor.

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