posted by Taylor on .
A particle has a charge of +1.5 µC and moves from point A to point B, a distance of 0.19 m. The particle experiences a constant electric force, and its motion is along the line of action of the force. The difference between the particle's electric potential energy at A and B is EPEA − EPEB = +5.50 10-4 J.
(a) Find the magnitude and direction of the electric force that acts on the particle.
(b) Find the magnitude and direction of the electric field that the particle experiences.