2.11 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?

Answers (1)

Given 

C=600pF

V=200V

Energy stored :

E=\frac{1}{2}CV^2=\frac{1}{2}*600*10^{-12}*200*200=1.2*10^{-5}J

Now, when it is disconnected and connected from another capacitor of capacitance 600pF

New Capacitance 

C'=\frac{600*600}{600+600}=300pF

New electrostatic energy

E'=\frac{1}{2}C'V^2=\frac{1}{2}*300*10^{-12}*200^2=0.6*10^{-5}J

Hence Loss in energy

E-E'=1.2*10^{-5}-0.6*10^{-5}J=0.6*10^{-5}

 

 

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