Question: A charge of +10μC is placed in a uniform electric field of strength 500 N/C. What is the work done in moving the charge 0.1m in the direction of the field?
Options:
Correct Answer: 0.5 J
Solution:
Work done W = F * d = (E * q) * d = (500 N/C * 10 × 10^-6 C) * 0.1 m = 0.5 J.