A particle has a charge of +1.6 µC and moves from point A to point B, a distance

Question

A particle has a charge of +1.6 µC and moves from point A to point B, a distance of 0.27 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 = +11 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.

I think I figured out (b). I did
E= 11X 10-4/ (1.6X10-6)(0.27m)
E= 11X10-4/4.32X10-7
E=2.54629X10-11
But, this answer still seems wrong.

For (a) I did:
E=F/q
I got 0.0040736N
Not sure if this is right.
Can someone read my work and see where I went wrong and explain? Thanks to anyone who helps!

Explanation / Answer

a) electric force=11 x 10-4 J*/0.27 =4.07*10^-3 N b) field= force/charge=4.07*10^-3/ +1.6 µC =2.5*10^3

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