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, C1 = 600 pF = 600 x 10-12 F
V1 = 200 V
Energy stored in the capacitor,
U1 = (1/2) C1 (V1)2 = (1/2) × 600 × 10-12 × (200)2
= 12 × 10-6 J
When this charged capacitor is connected to another uncharged capacitor C2 ( = 600 pF) ,they will share charges, till potential differences across their plates become equal.
Total charge on the two capacitors,
q = C1V1 + C2V2 = 600 × 10-12 × 200 + 0
= 12 × 10-8 C
Total capacitance of the two capacitors,
C = C1 + C2 = 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,
U2 = (1/2)CV2 = (1/2) x 1200 x 10-12 x (100)2
= 6 x 10-6 J
Therefore, energy lost by the capacitor C1( in the form of heat and electromagnetic radiation),
U1 – U2 = 12 x 10-6 – 6 x 10-6 = 6 x 10-6 J
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