Question: A charge of +3μC is placed in a uniform electric field of strength 1500 N/C. What is the work done in moving the charge 0.2 m in the direction of the field?
Options:
Correct Answer: 60 J
Solution:
Work done W = F * d = (E * q) * d = (1500 N/C * 3 × 10^-6 C) * 0.2 m = 0.0009 J = 90 J.