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)

Options

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`

Answer

B

Explanation

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))]`