A 4uf capacitor is charged by a 200V supply. The supply is then disconnected and the charged capacitor is connected to another uncharged 2 uf capacitor. How much electrostatic energy of the first capacitor is lost in the process of attaining the steady situation?

first we find the charge of the first capacitor by Q=CV, where V=200v and C=4(mu)f, then, we find the carge on the second capacitor, as the final voltage is same Q1/C1=Q2/C2; the we find the energy initially of the first capacitor and the the final energies of the first and second capacitors. E=Q^2/2C we subtract the initial from the final energy, we get the ans

Given:C1=4uf and V=200VThen,Charge in capacitor,Q=C1V=4×10^(-12)×200C=8×10^(-4).Now,C2=2ufThus common potential,V`=(C1V1+C2V2)/C1+C2=(4×10^(-6)×200)/6×10^-6=400/3Hence,heat energy lost=QV`/2=4×10^-4×400/6=8×10^-2/3=2.67×10^-2J