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
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Correct Answer: Option 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
Workdone = product of the charge Q, and the p.d V.
i.e Work done = QV
= 600 * 10-6 *2.1 * 103
=1.8j