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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 ? |
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Answer» Solution :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`. |
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