A 60 p F capacitor is fully charged by a 20 V supply. It is then disconnected from the supply and is connected to another uncharged 60 p F capacitor in parallel. The electrostatic energy that is lost in this process by the time the charge is redistributed between them is (in n J  )

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5 months ago

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Sol. Common potential after connection.

V common   = C 1 V 1 + C 2 V 2 C 1 + C 2 = 60 * 20 + 0 120 = 10 V o l t

 

Loss of energy = 1 2 C V 2 - 1 2 ( 2 C ) * V Common   2

= 1 2 * 60 * 10 - 12 * ( 20 ) 2 - 60 * 10 - 12 * ( 10 ) 2 = 60 * 10 - 12 ( 200 - 100 ) = 6000 * 10 - 12 = 6 n J

 

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Physics Electrostatic Potential and Capacitance 2025

Physics Electrostatic Potential and Capacitance 2025

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