A capacitor is charged to potential `V_(1)`. The power supply id disconnected and capacitor is connected in parallel to another uncharged capacitor. Calculate common potential of the combination of capacitors. Show that total energy of the combination is less than sum of energies stored in them before they are connected.

Explanation

Let `C_(1) and C_(2)` be the capacitances of the two capacitors and V be their common potential .
As `V = ("total Charge")/("total C apacity") = (q_(1) + q_(2))/(C_(1) + C_(2))`
`V = (C_(1) V_(1) + C_(2) V_(2))/(C_(1) + C_(2)) = (C_(1) V_(1))/(C_(1) + C_(2)) ( :' V_(2) = 0)`
Initial energy stored stored, `U_(1) = (1)/(2) C_(1) V_(1)^(2)`
Final energy after connection, `U_(2) = (1)/(2) (C_(1) + C_(2)) V^(2)`
`U_(2) = (1)/(2) (C_(1) + C_(2)) ((C_(1) V_(1))^(2))/((C_(1) + C_(2))^(2))`
`= (1)/(2) (C_(1)^(2) V_(1)^(2))/(C_(1) + C_(2)) = (U_(1) C_(1))/(C_(1) + C_(2))`
Clearly, `U_(2) lt U_(1)`.
which was to be proved.