Given, C1 = 600 pF = 600 x 10-12 FV1 = 200 VEnergy stored in the capacitor,U1 = (1/2) C1 (V1)2 = (1/2) × 600 × 10-12 × (200)2= 12 × 10-6 JWhen 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 CTotal capacitance of the two capacitors,C = C1 + C2 = 600 pF + 600 pF= 1200 pF= 1200 x 10-12 FTherefore, 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 JTherefore, 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

Q. 114.2( 77 Votes )

A 600pF capacitor

Class 12th NCERT - Physics Part-I 2. Electrostatic Potential and Capacitance

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,