Q. The charge on the capacitor of capacitance $C$ shown in the figure below will be
A) $CE$
B) $\frac{CER_{1}}{R_{1} + r}$
C) $\frac{CER_{2}}{R_{2} + r}$
D) $\frac{CER_{2}}{R_{1} + r}$
Explanation
(c) $I = \frac{R}{R_{2} + r}$
(since finally no current flows through capacitor)
∴ Potential difference across
$R_{2},V = IR_{2} = \frac{ER_{2}}{R_{2} + r}$
∴ Charge on the capacitor
$\mathcal{Q =}CV = \frac{CER_{2}}{R_{2} + r}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.