JAMB - Physics (2007 - No. 42)
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
Explication
Workdone = product of the charge Q, and the p.d V.
i.e Work done = QV
= 600 * 10-6 *2.1 * 103
=1.8j
i.e Work done = QV
= 600 * 10-6 *2.1 * 103
=1.8j