Question: A charge of +4 µC is placed in a uniform electric field of strength 500 N/C. What is the work done in moving the charge 0.2 m in the direction of the field?
Options:
Correct Answer: 200 J
Solution:
W = F * d = (E * q) * d = (500 N/C * 4 × 10^-6 C) * 0.2 m = 0.4 J.