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For section AB of a circuit shown in figure, $C_{1} = 1 μ F, C_{2} = 2 μ F, E = 10 V,$ and the potential difference $V_{A} - V_{B} = - 10 V .$ Charge on capacitor $C_{1}$ is

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$20 / 3 μ C$

$0 μ C$

none of these

$- 40 / 3 μ C$

answer is C.

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Detailed Solution

$V_{A} - \frac{q}{C_{1}} - E - \frac{q}{C_{2}} = V_{B} o r V_{A} - V_{B} - E = q (\frac{1}{C_{1}} + \frac{1}{C_{2}})$

or $q = - \frac{40}{3} μ C$

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