Work on done on a test charge

[rgo]

1. The problem statement, all variables and given/known data
I am looking at quation 10 but it relates to question 9
http://137.186.166.185:8080/question.gif

2. Relevant equations

In question 9 I calculated the electric feild to be 5.4 X10^6 N/c.
I did this by using the equation E = K(Q/r^2)


3. The attempt at a solution

I would assume that I would use this equation for question 10
W = PE2 - PE1 = -qEd

therfore

W = -(1.5x10^-6 C)(5.4 x 10^6 N/c)(0.25m)
W= -2.025 J

The sign is negative because the work is in the opposite direction of the electrical field.

This seems to be too simple. Am I missing something?

Thanks for the help

[turdferguson]

Youre moving a positive test charge from lower potential to higher potential, so the work done on the charge is potential energy at D - potential energy at B. What does that say about the sign?

[rgo]

Thats right, so my final fomula should be:
W = (1.5x10^-6 c)(5.4 X10^6 N/c)(0.25m) - (1.5x10^-6 c)(0)(0.25m)
W = 2.025 J

Thank you for your help

RG