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?
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Correct Answer: Option 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.
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.