Work Energy Problem Homework: Find v'B, W & in eV

[Mirth]

Homework Statement

A particle has a mass of m = 1.8*10^-5 kg and a charge of q = +3.25*10^-5 C. It is released from point A with an initial speed of v'A (v of A) = 3.5 m/sec and accelerates until it reaches point B. The charged particle moves along a straight line and does not rotate. The force on the particle is just the electric force. The potential difference between points A and B is V'A-V'B= 32.5 V.

a) Use ethe energy method to find the translational speed v'B of the particle at point B.
b) Find the work done by the electric force in Joules.
c) Convert the answer in part B into eV.

Homework Equations

Not sure.

The Attempt at a Solution

So after about an hour of hard work, I'm pretty sure that the answer is v'B = 11.38 m/sec.

I have NO clue what formula I would use to find Work for question b. The closest thing I can find in my book for this chapter is EPE'B - EPE'A = -W'AB. And from other chapter, W = F*d.

If anyone could help me get on track, I'd greatttly appreciate it.

 

[cepheid]

The closest thing I can find in my book for this chapter is EPE'B - EPE'A = -W'AB. And from other chapter, W = F*d.

Yes exactly. The change in potential energy of the particle is equal to the negative of the work done on it between points A and B. So, what is the change in the particle's potential energy? Hint: how is it related to the potential difference between those two points?