00:01
In this problem, a particle with charge q, or in this case positive 1 .5 microculems, moves from a to b, which are a distance d apart, or 0 .20 meters, and under the effect of a constant magnitude electric force f.
00:25
We're asked to find the magnitude of that force f, as well as the magnitude and direction of the electric field that the particle experiences.
00:34
We're also given that the difference between the electric potential energy at a and the electric potential energy at b is equal to 9 .0 times 10 to the minus 4 joules.
00:51
We also know that the difference between potential energy from a to b is the work done on the particle.
01:01
We also know that work is equal to a force over a distance.
01:11
This force happens to be the electro -electric force that acts on the particle over that distance d.
01:20
We can now begin to solve for the magnitude of that force.
01:26
Thus, the change in electric potential energy from a to b over the distance between a to b, equals the magnitude of the force...