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Heating Effect of Electric Current

The charges o...

Question

The charges on the capacitor of capacitance C at steady state shown in fig will be:

A

C E

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B

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

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C

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

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D

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

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Solution

The correct option is C $\frac{C E R_{2}}{R_{2} + r}$
Since the potential difference between $a$ to $d$ and $b$ to $c$ is equal.

It can be written as,

$V_{a d} = V_{b c}$

$V_{a d} = (\frac{E}{r + R_{2}} \times R_{2})$

The charge on the capacitor is given as,

$Q = C V_{a d}$

$= (\frac{C E R_{2}}{r + R_{2}})$

Thus, the charge on the capacitor is $(\frac{C E R_{2}}{r + R_{2}})$ .

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