A capacitor of capacitance `C_(1) = 1.0 muF` carrying initially a voltage `V = 300 V` is connected in parallel with an uncharged capacitor of capacitance `C_(2) = 2.0 muF`. Find the increment of the electric energy of this system by the moment equilibrium is reached. Explain the result obtained.

Explanation

Charge contained in the capacitor fo capacitance n`C_(1)` is `q = C_(1) varphi` and the energy, stored in it :
`U_(i) = (q^(2))/(2 C_(1)) = (1)/(2) C_(1) varphi^(2)`
Now, when the capacitor are connected in parallel, equivalent capacitance of the system, `C = C_(1) + C_(2)` and hence, energy stored in the system :
`U_(f) = (C_(1)^(2) varphi^(2))/(2 (C_(1) + C_(2)))`, as charge remains conserved during the process.
So, increment in the energy,
`Delta U = (C_(1)^(2) varphi^(2))/(2) ((1)/(C_(1) + C_(2)) - (1)/(C_(1))) = (-C_(2) C_(1) varphi^(2))/(2 (C_(1) + C_(2))) = 0.03 mJ`