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Two identical parallel plate capacitors ...

Two identical parallel plate capacitors of capacitance C each are connected in series with a battery of emf, E as shown. If one of the capacitors is now filled with a dielectric of dielectric constant k, the amount of charge which will flow through the battery is (neglect internal resistance of the battery)

A

`(k+1)/(2(k-1))CE`

B

`(k-1)/(2(k+1))CE`

C

`(k-2)/((k+1))CE`

D

`(k+2)/((k-1))CE`

Text Solution

Verified by Experts

The correct Answer is:
B
Initial charge on both capacitor
`Q_i = (CE)/2`

Now, charge on each capacitor
`Q_f = ((KC)/(K+1))E `

Change in charge on C is supplied by battery
Charge supply by battery `=((KC)/(K+1))E-(CE)/2`
`=CE [(K-1)/(2(K+1))]`
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