If two charged plates are maintained at a potential difference of 3 kv, the work done in taking a charge of 600 µC across the field is?
- 0.8 J
-
1.8 J
- 9.0 J
- 18.0 J
The correct answer is: B
Explanation
Work done = product of the charge Q, and the p.d V.(i)/e work done = Q.V = 600 x 10-6 x 3.0 x 10-3 = 1.8 J.