Question

The charge on the capacitor of capacitance $C$ shown in the figure below will be

• A. $C\,E$
• B. $\frac{CE\,R_{1}}{R_{1}+r}$
• C. $\frac{CE\,R_{2}}{R_{2}+r}$
• D. $\frac{CE\,R_{1}}{R_{2}+r}$

Answer 1

Answer: (C)

$\frac{CE\,R_{2}}{R_{2}+r}$

Answer 2

From the figure, we see that the current in the circuit would be equal to
$I\left(R_{2}+r\right)=E$
As no current flows through the capacitance in a DC connection.
Thus, we get $I=\frac{E}{\left(R_{2}+r\right)}$
Thus, the potential across the resistance $R_{2}$ is equal to
$V=I R_{2}=\frac{E R_{2}}{\left(R_{2}+r\right)}$
and thus the charge on the capacitor would be equal to
$Q=C V=\frac{E R_{2} V}{\left(R_{2}+r\right)}$