Q. In the given figure each plate of capacitance $C$ has partial value of charge
A) $CE$
B) $\frac{CER_{1}}{R_{2} - r}$
C) $\frac{CER_{2}}{R_{2} + r}$
D) $\frac{CER_{1}}{R_{1} - r}$
Explanation
(c) In steady state current drawn from the battery
$i = \frac{E}{(R_{2} + r)}$
In steady state capacitor is fully charged hence No current will flow through line (2)
Hence potential difference across line (1) is $V = \frac{E}{(R_{2} + r)} \times R_{2},$ the same potential difference appears across the capacitor, so charge on capacitor
$Q = C \times \frac{ER_{2}}{(R_{2} + r)}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.