Question

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

• A. $CE$
• B. $\frac{CE{R}_1}{R_2+r}$
• C. $\frac{CE{R}_2}{R_2+r}$
• D. $\frac{CE{R}_1}{R_1+r}$

Answer 1

The correct option is D $\frac{CE{R}_1}{R_1+r}$

At steady-state, no current flows through the capacitor. Thus a capacitor behaves like an open switch.
Equivalent circuit is shown in the figure.

Effective resistance of the equivalent circuit, $R_{eff}=R_1+r$
Now, Potential drop across resistance $R_1,V=\frac{I}{R_1}=\frac{R_1}{R_{eff}}E$
$\therefore V=\frac{R_1}{R_1+r}E$

As $R_1$ and $C$ are connected in parallel to each other. Thus potential drop across $C$ is same as that across $R_1$.

Thus charge on the capacitor plates, $Q=CV$
$\therefore Q=\frac{R_{1}CE}{R_1+r}$