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

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Q. The charge on the capacitor of capacitance C shown in the figure below will bePhysics Question Image

Electromagnetic Induction

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
$V=IR_2=\frac{ER_2}{(R_2+r)}$and thus the charge on the capacitor
would be equal to
$ \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, Q=CV=\frac{ER_2V}{(R_2+r)}$