A 600pF capacitor is charged by a 200V supply. It is then disconnected from the supply and is connected to another uncharged 600 pF capacitor. How much electrostatic energy is lost in the process? (1 pF = 10−12 F)
Energy stored in a capacitor :-
E=12CV2
E=1.2×10−5J
Another capacitor is connected after disconnecting the battery.
Equivalent capacitance (C') is;
1C′=1C+1C
⇒C′=300pF
New Energy, E=12CV2
⇒E′=0.6×10−5J
Loss in energy, ΔE=E′−E=6×10−5J