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A 600 pF capacitor is charged by a 200 V supply. It is then disconnected from the supply and is connected to another uncharged 600 pF capacitor. What is the common potential in V and energy lost in J afrte reconnection?

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Answer Text

Solution :
Capacitance of the capacitor, C=600 pF <br> potential diffference V=200 V <br> Elec trostatic energy stored in the capacitor is given by, <br> `E=1/2 CV^(2)` <br> `=1/2 xx(600xx10^(-12)) xx(200)^(2)` <br> =`1.2xx10^(-5) J` <br> If supply is disconnected from the capacitor and another capacitor of capacitance C=600 pF is connected to it, then equivalent capacitance (C') of the combination is given by , `1/(C')=1/C+1/C` <br> `=1/600+1/600=2/600 =1/300` <br> `:. C'=300 pF` <br> New electrostatic energy can be calculated as <br> `E'=1/2xxC'xxV^(2)` <br> =`1/2xx300xx(200)^(2)` <br> `=0.6xx10^(-5)J` <br> loss in electrostatic energy =E-E' <br> `=1.2xx10^(-5)-0.16xx10^(-5)` <br> `=0.6xx10^(-5)` <br> `=6xx10^(-6) J` <br> Therefore, the electrostatic energy lost in the process is `6xx10^(-6)` J

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