Is the Work Done by the Electric Field Positive or Negative?

[tomwilliam2]

Homework Statement

How much work is done by the electric field in moving a particle from (a,a,0) to (a,a,a) in a region where the electric field is:
E = zye_x + yxey + xyez

Homework Equations

F=qE
W = integral F dot dl
V(2)-V(1)= - integral E dot dl

The Attempt at a Solution

I know how to do this, just holding x=a, y=a and integrating from z=0 to z=a.
I get the answer W = qa³
The trouble is, the other students get W=-qa³
If I do this as the line integral of the electric field times the charge, I get positive work (and I'm after the work done by the field, not some external agent), but if you do it as the change in potential it comes out negative. So which is correct? I'm drawing an analogy with a gravitational field, in which I'd have to do negative work (as an external agent) to move an object in the direction of the field, so presumably the field itself is doing positive work?
Need help...exam on Tuesday!
Thanks in advance

 

[Hypersphere]

It depends in which order you take the points in the potential difference. V(2)-V(1) would be the difference in energy if you go from point two to point one, while you consider the opposite path and thus gets the opposite sign.

 

[tomwilliam2]

The questions states that the particle is moving from (a,a,0) to (a,a,a), so moving in the positive z-direction. When you say "difference in energy" that doesn't make it clear to me how I should answer the question. The question wants the work done BY the electric field. I claim that if the field exerts a force moving a positively charged particle through a positive path in the direction of the force, then the field is doing positive work, while the force required (by an external agent) to move the particle along this same path would be negative.