# 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

Numerical

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

= 1/2 xx (600 xx 10^-12) xx (200)^2

= 1.2 xx 10^-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 = 2/600 = 1/300

C' = 300 pF

New electrostatic energy can be calculated as

"E'" = 1/2 xx "C'" xx "V"^2

= 1/2 xx 300 xx (200)^2

= 0.6 xx 10^-5  "J"

Loss in electrostatic energy = E − E'

= 1.2 xx 10^-5 - 0.6 xx 10^-5

= 0.6 xx 10^-5

= 6 xx 10^-6  "J"

Therefore, the electrostatic energy lost in the process is 6 xx 10^-6  "J".

Concept: Potential Due to a System of Charges
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#### APPEARS IN

NCERT Physics Part 1 and 2 Class 12
Chapter 2 Electrostatic Potential and Capacitance
Exercise | Q 2.11 | Page 87
NCERT Class 12 Physics Textbook
Chapter 2 Electrostatic Potential and Capacitance
Exercise | Q 11 | Page 88

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