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Question: An air capacitor of capacity \(C = 10\mu F\)is connected to a constant voltage battery of\(12V\). No...

An air capacitor of capacity C=10μFC = 10\mu Fis connected to a constant voltage battery of12V12V. 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

A

120μC120\mu C

B

699μC699\mu C

C

480μC480\mu C

D

24μC24\mu C

Answer

480μC480\mu C

Explanation

Solution

Initially charge on the capacitor

Q=10×12=120μCQ = 10 \times 12 = 120\mu C

Finally charge on the capacitor

Q′=(5×10)×12=600μCQ' = (5 \times 10) \times 12 = 600\mu C

So charge supplied by the battery later =Q′−Q=480μC= Q' - Q = 480\mu C