A capacitor having a capacitance of `100muF` is charged to a potential difference of `24V` . The charging battery is disconnected and the capacitor is connected to another battery of emf `12V` with the positive plate of the capacitor joined with the positive terminal of the battery . (a ) Find the charges on the capacitor before and after the reconnection . (b ) Find the charge flown through the 12V battery . (c ) Is work done by the battery or is it done on the battery ? find its magnitude . (d ) Find the decrease is electrostatic field energy . (e ) Find the best developed during the flow of charge after re-connection.

Options

A. 21.6mJ
B. 27.6mJ
C. 7.2mJ
D. zero

Answer

C

Explanation

Work done by battery, `W=V(Deltaq)=12(-1200xx10^(-6))=-14.4mJ`.
Now `U_(i)+W=U_(f)+`heat generated or `28.8-14.4=7.2+H" "therefore H=7.2mJ`.