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 2m in the direction of the field?
Practice Questions
1 question
Q1
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 2m in the direction of the field?
10 J
1 J
100 J
0.5 J
Work done W = F * d = (E * q) * d = (500 N/C * 10 × 10^-6 C) * 2m = 0.01 J.
Questions & Step-by-step Solutions
1 item
Q
Q: 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 2m in the direction of the field?
Solution: Work done W = F * d = (E * q) * d = (500 N/C * 10 × 10^-6 C) * 2m = 0.01 J.
