A capacitor `A` of capacitance `4muF` is charged to `30V` and another capacitor `B` of capacirtance `muF` is charged to `15V`. Now, the opositive plate of `A` is connected to the negative plate of `B` and negative plate of `A` to the positive plate of `B`. find the final charge of each capacitor and loss of electrostatic energy in the process.

Explanation

Here charge on `A, Q_(1)=C_(1)=C_(1)V_(1)=4xx30=120muC`
Charge on `B, Q_(2)=C_(2)V_(2)=2xx15=30muC`
When `+ve` plate is connected to `-ve` plate, we find net charge as `|Q_(1)-Q_(2)|=120-30=90muC`.
This `90muC` cahrge will be redistributed in such a manner that capacitors acquire same potential.
Common potential:
`V=(|C_(1)V_(1)-C_(2)V_(2)|)/(C_(1)+C_(2))=(4xx30-2xx15)/(4+2)=15V`
`Q'_(1)=C_(1)V=4xx15=60muC`
`Q'_(2)=C_(2)V=2xx15=30muC`
Loss of energy:
`DeltaU=(1)/(2)(C_(1)C_(2))/(C_(1)+C_(2))(V_(1)+V_(2))^(20`
`=(1)/(2).(4xx2)/(4+2)(30-15)xx10^(-6)J`
`=150muJ`