Question

A capacitor is charged by a battery. The battery is removed and another identical uncharged capacitor is connected in parallel. The total electrostatic energy of resulting system:
A. increasese by a factor of 4
B. decreases by factor of 2
C. remains the same
D. increases by a factor of 2

Answer 1

Correct Answer - D
Tahinking Process Energy stored in a system of capacitors `=sum (1)/(2)CV^(2)`
Also,potential drop remains same in parallel across both capacitors
Inteially stored energy
" "`U_(1)=(1)/(2)CV^(2)`
Finally,potential drop across each capacitor will be still V.
so, finally stord energy
`U_(2)-(1)/(2)CV^(2)+(1)/(2)cV^(2)`
`=(1)/(2)(2C)V^(2)=2((1)/(2)CV^(2))=2U_(1)`