Question

The potential at a point A in an electric field is 300 volt and at a point B it is 1200 volt. What work must be done to move a positive charge of 3x10^-8 coulomb from A to B?​

Answer 1

Explanation:

The energy stored on a capacitor can be calculated from the equivalent expressions: This energy is stored in the electric field. and will have stored energy E = x10^ J. From the definition of voltage as the energy per unit charge, one might expect that the energy stored on this ideal capacitor would be just QV.

Answer 2

Answer:

VA = 300V

VB = 1200V

q = 3 x 10^-8C

Work done = VB - VA / q = 1200 - 300 / 3 x 10^-8 = 3 x 10^10 J

HOPE IT HELPS... ✌