A 4 `muF` capacitor is charged by a 200 V supply. It is then disconnected from the supply, and is connected to another uncharged `2 muF` capacitor. How much electrostatic energy of the first capacitor is lost in the form of heat and electromagnetic radiation ?

Explanation

Here `C_1 = 4 muF = 4 xx 10^(-6)F, V_1 = 200 V, C_2= 2muF = 2 xx10^(-6)F and V_2 = 0`
On sharing of charges loss of electrostatic energy
`Delta u = (C_1C_2(V_1-V_2)^2)/(2(C_1 + C_2)) = (4 xx 10^(-6) xx 2 10^(-6) xx (200-0)^2)/(2(4 xx 10^(-6) + 2 xx 10^(-6))`
`=(4 xx 10^(-6) xx 2 xx 10^(-6) xx 200 xx 200)/(2 xx 6 xx 10^(-6)) = 2.67 xx 10^(-2)J`.