Consider two uncharged capacitors of equal capacitance 200 pF. One of them is charged by a 100 V supply and disconnected. Now this capacitor is connected to the uncharged capacitor. The amount of electrostatic energy lost in the process is :
- 0.5 J
- 1.0 10-6 J
- 0.5 10-6 J
- 1.0 J
Answer
(3) 0.5 10-6 J
Solution
Before: Ui = 12CV2 = 12 200 10-12 (100)2 = 1.0 10-6 J
Charge q = CV = 200 10-12 100 = 2 10-8 C is shared equally ⇒ common potential V' = V2 = 50 V
After: Uf = 2 12 200 10-12 (50)2 = 0.5 10-6 J
∴ Energy lost = Ui - Uf = 0.5 10-6 J (lost as heat and radiation)
Answer: (3)