An air capacitor of capacity C=10μF 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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120μC

699μC

480μC

24μC

answer is C.

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Detailed Solution

Initially charge on the capacitor Qi=CoV=10×12=120μCFinally charge on the capacitor Qf=CV=KCoV=5×10×12=600μCSo charge supplied by the battery later =Qf−Qi=480μC.

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