Two capacitors A and B of capacities C and 2C are connected in parallel and the combination is connected to a battery of V volts. After the charging is over , the battery is removed. Now a dielectric slab of `K=2` is inserted between the plates of A so as fill the completely. The energy lost by the system during the sharing of charges is
(ii) Without battery of parallel combination :

Explanation

`C_(p)^(1)=KC+2C=2C+2C=4C" [because K=2]`
`V^(1)=(q)/(C_(p)^(1))=(3CV)/(4C)=(3V)/(4)`
Final Energy stored, `U_(f)=(1)/(2)C_(p)^(1)(V^1)^2`
`U_(f)=(1)/(2)xx4Cxx((3V)/(4))^2`

`therefore U_(f)=(1)/(2)xx4Cxx(9v^2)/(16)=(9CV^2)/(8)`
`therefore` The energy lost by the system
`Delta U=U_(1)=(3)/(2)CV^(2)-(9CV^2)/(8)`
`therefore Delta U=(3CV^2)/(8)` .