198k views
5 votes
A 3.0-mC point charge and a 10.0-mC point charge are initially extremely far apart. How much work does it take to bring the 3.0-mC point charge to a position 6.0 mm frin tge 10.0-mC point charge?

A. 90 J
B. 30 J
C. 7.5 J
D. 45 J

User Sam Bevins
by
8.8k points

1 Answer

4 votes

Final answer:

The work done in bringing a point charge from one point to another is given by the equation: W = q(Vf - Vi). In this case, we have a 3.0-mC point charge and a 10.0-mC point charge. Initially, the charges are extremely far apart, so the initial potential is considered negligible. Bringing the 3.0-mC point charge to a position 6.0 mm from the 10.0-mC charge means the final potential is non-zero.

Step-by-step explanation:

The work done in bringing a point charge from one point to another is given by the equation:

W = q(Vf - Vi)

where W is the work done, q is the charge, and Vf and Vi are the final and initial potentials, respectively.

In this case, we have a 3.0-mC point charge and a 10.0-mC point charge. Initially, the charges are extremely far apart, so the initial potential is considered negligible. Bringing the 3.0-mC point charge to a position 6.0 mm from the 10.0-mC charge means the final potential is non-zero. Therefore, the work done can be calculated using the equation above with the given values.

User GregoryComer
by
8.8k points
Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.