# 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?

#### 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