Question: A charge of +5 µC is placed in an electric field of 300 N/C. What is the work done in moving the charge 0.2 m in the direction of the field? (2019)
Options:
Correct Answer: 6 J
Exam Year: 2019
Solution:
Work done W = F * d = (E * q) * d = (300 N/C * 5 × 10^-6 C) * 0.2 m = 0.3 J.