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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. How Much Electrostatic Energy is Lost in the Process? - Physics

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. How much electrostatic energy is lost in the process?

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Solution

Capacitance of the capacitor, C = 600 pF

Potential difference, V = 200 V

Electrostatic energy stored in the capacitor is given by,

`E=1/2 CV^2`

`=1/2xx(600xx10^-12)xx(200)^2`

`=1.2xx10^-5 J`

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`

`=1/600+1/600=1/300`

`C'=300 pF`

New electrostatic energy can be calculated as

`E'=1/2xxC'xxV^2`

`=1/2xx300xx(200)^2`

`=0.6xx10^-5 J`

Loss in electrostatic energy=E-E'

`=1.2xx10^-5-0.6xx10^-5`

`=0.6xx10^-5`

`=6xx10^-6 J`

Therefore, the electrostatic energy lost in the process is `6xx10^-6 J`.

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APPEARS IN

NCERT Class 12 Physics Textbook
Chapter 2 Electrostatic Potential and Capacitance
Q 11 | Page 88
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