An air capacitor of capacity C=10 $μF$ is connected to a constant voltage battery of 12 volt. Now the space between the plates is filled with a liquid of dielectric constant 5. The (additional) charge that flows now from battery to the capacitor is

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$600 μC$

$24 μC$

$480 μC$

$120 μC$

answer is C.

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Detailed Solution

$\begin{array}{l} q_{1} = C_{1} V = 10 \times 12 = 120 μC \\ q_{2} = C_{2} V = {KC}_{1} \times V = 5 \times 10 \times 12 = 600 μC \end{array}$
Additional charge that flows
$= q_{2} - q_{1} = 600 - 120 = 480 μC$

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