A particle has a charge of +1.7 μC and moves from point A to point B, a distance of 0.26 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 = +8.1 x 10-4 J. (a) Find the magnitude of the electric force that acts on the particle. (b) Find the magnitude of the electric field that the particle experiences.
The Attempt at a Solution
What's wrong with this reasoning:
F = V/d and V=Ed so E=F/q=V/d
substitute all those values into that formula to get the electrostatic force and it's wrong?
I also tried b) using E=kq/d^2 and that didn't work either! (I got 226331.3609 as the answer)
By the way, there are no rounding requirements or anything, only a 2% margin of error so it's not that.
Thanks a lot for your help in advance!