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Partially Filled Dielectrics

The charge on...

Question

The charge on the capacitor of capacitance $C$ shown in the figure will be :

C E

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\frac{C E R_{1}}{R_{1} + r}

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\frac{C E R_{2}}{R_{2} + r}

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\frac{C E R_{1}}{R_{2} + r}

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Solution

The correct option is D $\frac{C E R_{2}}{R_{2} + r}$
The direct current $(I)$ is flowing through the circuit, thus the capacitor will not allow the DC to pass through it, i.e., the current will flow through the resistance $r$ and $R_{2}$ only.
$⟹$ $E = I (R_{2} + r)$
$⟹ I = \frac{E}{(R_{2} + r)}$ ........(1)

Potential difference between point A and B:
$\Rightarrow V_{A B} = I R_{2} = \frac{E R_{2}}{(R_{2} + r)}$

As the current through $R_{1}$ is zero, thus, $V_{P Q} = V_{A B}$
$\Rightarrow$ $V_{P Q} = \frac{E R_{2}}{(r + R_{2})}$

Let the charge on the capacitor be $q$ :
$\Rightarrow$ $q = C V_{P Q}$
$\Rightarrow q = \frac{C E R_{2}}{(r + R_{2})}$

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