Two identical capacitors, have the same capacitance C. One of them is charged to potential `V_1` and the other `V_2`. The negative ends of the capacitors are connected together. When the poistive ends are also connected, the decrease in energy of the combined system is

Options

A. `(1)/(4)C(V_(1)^(2) - V_(2)^(2))`
B. `(1)/(4)C(V_(1)^(2) + V_(2)^(2))`
C. `(1)/(4)C(V_(1) - V_(2))^(2)`
D. `(1)/(4)C(V_(1) +V_(2))^(2)`

Answer

C

Explanation

The initial energy of the system,
`U_(i) = (1)/(2)CV_(1)^(2) + (1)/(2)CV_(2)^(2) = (1)/(2)C (V_(1)^(2) + V_(2)^(2))`
When the capacitors are joined, they reach a common potential V.
`V = (q_(1) + q_(2))/(C_(1) + C_(2)) = (CV_(1) + CV_(2))/(C + C) = (V_(1) + V_(2))/(2)`
The final energy of the system,
`U_(f) = (1)/(2)(C + C)V^(2) = (1)/(2)(2C)V^(2)`
Decrease in energy,
`U_(i) - U_(f) = (1)/(2)C(V_(1)^(2) + V_(2)^(2)) - (1)/(2)(2C) ((V_(1) + V_(2))/(2))^(2)`
`= (1)/(4)C(V_(1) - V_(2))^(2)`