If two charged plates are maintained at a potential difference of 3 kv, the work done in taking a charge of 600µc across the field is
- 0.8j
-
1.8j
- 9.0j
- 18.0j
The correct answer is: B
Explanation
Workdone = product of the charge Q, and the p.d V.i.e Work done = QV
= 600 * 10-6 *2.1 * 103
=1.8j