Question

A battery of 10V is connected to a capacitor of capacitance 0.1F. The battery is now removed from the battery and this capacitor is connected to a second uncharged capacitor, find the total energy stored in each capacitor, and compare with the initial energy of the first capacitor.

Answer 1

Answer:

Capacitance of the capacitor, C= 0.1 F

Potential applied across the capactor, V = 10 V

Therefore,

Energy stored in the first capacitor is

Ui = 12CV2 = 12×0.1×(10)2 = 5 J.

Given, that the battery is removed and the charged capacitor is connected to an uncharged capacitor of the same capacitance.

Suppose the common potential is V.

Then, using the law of conservation of charge,

Charge on each capacitor, q' = CV'

and, q' = q2

∴ Total energy stored in the capacitor is

Uf = 2 × 12q'V' = q'×q'C = 14q2C q' = q/2, q=CV

= 12×12CV2=12× Ui= 12×5 = 2.5 J.