An isolated 510 F parallelplate capacitor has 410 mC of char

An isolated 5.10 F parallel-plate capacitor has 4.10 mC of charge. An external force changes the distance between the electrodes until the capacitance is 2.10 F.

Part A How much work is done by the external force?

Solution

Ui = Q^2/2Ci = (4.1*10^-3)^2/(2*5.1*10^-6) = 1.64 J

Uf =   Q^2/2Cf = (4.1*10^-3)^2/(2*2.1*10^-6) = 4. J

worj = Uf - Ui = 4 - 1.64 = 2.36 J