Question

The electric potential at point a is 10 volt and b is -10 volt then the work done by external agent carrying the charge of 1 into 10 raise to minus 6 coulomb from A to B is

Answer 1

The work done by the external agent carrying the charge 10¯⁶ C from A to B is 2 × 10¯⁵ J.

The electric potential at point ‘A’ is 10 volt and ‘B’ is -10 volt.

We have to find the work done by external agent carrying the charge of 10¯⁶ C from A to B.

We know,

Work done by the external agent carrying the charge from A to B = potential difference from A to B × magnitude of charge.

= [10V - (-10V)] × 10¯⁶ C

= 20 × 10¯⁶ J

= 2 × 10¯⁵ J

Therefore the work done by the external agent carrying the charge 10¯⁶ C from A to B is 2 × 10¯⁵ J.