Change in potential / potential energy of a charge in electric field [closed]

Question

A uniform electric field of 2000 $(\frac V m)$ exists through all space (the field is in $+x$ direction). An electron is moved from the origin to the location where $x = -2.00 (m)$ and $y = +3.00 (m)$. What is the change in potential energy and change in electric potential respectively?

I calculated the electric potential energy change to equal $6.4\times10^{-16}$ J and the voltage to equal 4000V. This all is correct after confirmation with the problem hint. However, I am unsure whether the voltage and electric potential energy are negative or positive. I am sure they are both negative because the electron is moving from a high energy area to a low energy area because it's getting closer to the electric field source. However, this is wrong. I would really appreciate a thorough explanation as to what the correct sign is for the voltage and potential energy.

Answer 1

Though you correctly mentioned that the electron is moving from a high energy region to a low energy region, the conclusion that you made is not correct. The electric potential energy of the system will, indeed, decrease, and thus, the change in potential energy is negative. But the change in electric potential (a.k.a. 'voltage') will be positive since you are moving in a direction opposite to that of the electric field. The change in electric potential is independent of the charge (electron in this case) that is being moved. It depends only on the strength and the direction of the electric field, assuming that this field is not affected by the movement of the charge.