Q. 114.4( 51 Votes )

# A 600pF capacitor is charged by a 200V 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?

Answer :

Given, C_{1} = 600 pF = 600 x 10^{-12} F

V_{1} = 200 V

Energy stored in the capacitor,

U_{1} = (1/2) C_{1} (V_{1})^{2} = (1/2) × 600 × 10^{-12} × (200)^{2}

= 12 × 10^{-6} J

When this charged capacitor is connected to another uncharged capacitor C_{2} ( = 600 pF) ,they will share charges, till potential differences across their plates become equal.

Total charge on the two capacitors,

q = C_{1}V_{1} + C_{2}V_{2} = 600 × 10^{-12} × 200 + 0

= 12 × 10^{-8} C

Total capacitance of the two capacitors,

C = C_{1} + C_{2} = 600 pF + 600 pF

= 1200 pF

= 1200 x 10^{-12} F

Therefore, common potential of the two capacitors,

Energy stored in the combination of the two capacitors,

U_{2} = (1/2)CV^{2} = (1/2) x 1200 x 10^{-12} x (100)^{2}

= 6 x 10^{-6} J

Therefore, energy lost by the capacitor C_{1}( in the form of heat and electromagnetic radiation),

U_{1} – U_{2} = 12 x 10^{-6} – 6 x 10^{-6} = 6 x 10^{-6} J

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