The numerical value of the charge on either p | vedclass

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(D) In the steady state,the capacitor $C$ acts as an open circuit,meaning no current flows through the branch containing the capacitor.The circuit sim

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

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(D) In the steady state,the capacitor $C$ acts as an open circuit,meaning no current flows through the branch containing the capacitor.The circuit simplifies to a series combination of the battery $E$,internal resistance $r$,and resistor $R_2$.The current $i$ flowing through the circuit is given by:$i = \frac{E}{R_2 + r}$The potential difference across the resistor $R_2$ (between points $d$ and $c$) is:$V_{dc} = i R_2 = \frac{E R_2}{R_2 + r}$Since the capacitor $C$ is connected in parallel with the resistor $R_2$,the potential difference across the capacitor is equal to the potential difference across $R_2$.Therefore,the charge $Q$ on the capacitor is:$Q = C V_{dc} = \frac{C E R_2}{R_2 + r}$

The numerical value of the charge on either plate of the capacitor $C$ shown in the figure is

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