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# A 600 p F capacitor is charged by a 200 V supply. It is then disconnected from the supply and is connected to another uncharged 600 p F capacitor. How much electrostatic energy is lost in the process?

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?

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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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