(Very Basic) Movement of Charged Particles Through Parallel Plate Apparatus

[012anonymousx]

Homework Statement

What is the electric potential two-fifths of the way through a parallel-plate apparatus (from the positive plate) if the plates have a total separation of 5cm and field strength of 5000N/C.

Homework Equations

Difference Potential Energy = Electric Field Strength x Distance
V = ed

The Attempt at a Solution

= 5000 x (0.05 - ((2/5)(0.05))
= 5000 x (0.05-0.02)
= 5000 x (0.03)
= 150

Real answer is 250, so basically the whole distance of 0.05.
I thought that electric potential decreases as the current flows from positive to negative?
That is why, in my solution, I took the two-fifth of the whole distance away.
Edit: Please let me know if my solution as infact correct :)

 

[gneill]

Your solution looks fine to me. As you say, the total potential difference is 250V, so a correct answer for the potential partway between the plates can't be 250V also.