Question

A 500 pF capacitor is charged by a 200 V battery
(a) How much electrostatic energy is stored in the capacitor?
(b) if the capacitor is disconnected from the battery and connected in parallel to ot, find the electrostatic energy stored in the system
another 400 pF F capacitor,
(c) Why is there a difference in the two cases?​

Answer 1

Answer:

Explanation:

Capacitance of the capacitor, C = 600 pF

Potential difference, V = 200 V

Electrostatic energy stored in the capacitor is given by,

If supply is disconnected from the capacitor and another capacitor of capacitance C = 600 pF is connected to it, then equivalent capacitance (C') of the combination is given by,

New electrostatic energy can be calculated as

Loss in electrostatic enegy = E - E'

                                       = 1.2 x 10-5  -  0.6 x 10-5

                                       = 0.6 x 10-5

                                       = 6 x 10-6 J  

Therefore, the electrostatic energy lost in the process is 6 x 10-6 J.