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

2365 234 Electromagnetic Induction Report Error

A

CE

11%

B

$\frac{CE R _{1}}{R _{1} + r}$

29%

C

$\frac{CE R _{2}}{R _{2} + r}$

36%

D

$\frac{CE R _{1}}{R _{2} + r}$

25%

Solution:

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