Question

The numerical value of charge on either plate of capacitor C shown in figure:

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

Answer 1

Answer: (D)

$\frac{CE{R}_1}{R_1+r}$

At steady state, no current flows through the capacitor. Thus a capacitor behaves like a open switch.

Equilibrium circuit is shown in the above figure.

Effective resistance of the equivalent circuit $R_{eff}=R_1+r$

Potential drop across resistor $R_1$, $V=\frac{R_1}{R_{eff}}E$

$\therefore$ $V=\frac{R_1}{R_1+r}E$

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 $Q=CV$

$\therefore$ $Q=\frac{R_{1}CE}{R_1+r}$