`Q=CV_(0)`
They have equal potential difference, and thir combined charge must add up to the original charge Therefore,
`V=Q_(1)/C_(1)=Q_(2)/C_(2)` and also `Q_(1)+Q_(2)=CV_(0)`
`C_(1)=C` and `C_(2)=C/2`,
So `Q_(1)/C=Q_(2)/((C//2))`
or `Q_(2)=Q_(1)/Q_(2)`
or `Q=3/2Q_(1) or Q_(1)=2/3Q`
So `V=Q_(1)/C_(1)=2/3Q/C=2/3V_(0)`
c. `U=1/2(Q_(1)^(2)/C_(1)+Q_(2)^(2)/C_(2))=1/2[(2/3Q)^(2)/C+(2(1/3Q)^(2))/C] `
`=1/2Q_(2)/C=1/2CV_(0)^(2)`
d. The original `U` was `U=1/2CV_(0)^(2)` or `DeltaU=(-1)/6CV_(0)^(2)`.
Thermal energy of capacitor, wires, etc., and electromagnetic radiation.