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

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$120 μ C$

$699 μ C$

$24 μ C$

$480 μ C$

answer is C.

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Detailed Solution

Initially charge on the capacitor $Q_{i} = C_{o} V = 10 \times 12 = 120 μ C$
Finally charge on the capacitor $Q_{f} = C V = K C_{o} V = (5 \times 10) \times 12 = 600 μ C$
So charge supplied by the battery later $= Q_{f} - Q_{i} = 480 μ C$ .

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