Electric potential concept Imagine having two charged plates, one positive and one negative, and a negative point charge is placed at the negative plate. Let's set the negative plate to zero potential. The distance between the negative point charge and the negative plate is zero, so $V$ is zero in the equation $V=Ed$. However, since $U=qV$, potential energy is zero at this point. This should not be correct because the negative point charge of course has potential energy at this location. What is the conceptual error in this thought process?
 A: The value of the potential energy doesn't matter, just the change in potential energy. You can set any level of potential to be the zero point.  Think of gravitational energy $mgh$, what is $h$? You could measure it from sea level, or you could measure it from the center of the earth. If you're right at sea level, then if you measure your height from sea level your potential energy is 0 but if from the center of the earth you're potential energy is $mg R_{earth}$. The only thing you would really care about, was what you're speed would be if you feel down a $100m$ pit. In that case your kinetic energy would be equal to the CHANGE in potential energy, which would be $mg(100\  \text{meters})$ in both situatons.
In your case you're electron does have potential energy at the negative plate, it has 0 potential energy, but say the positive plate is at $+100$ Volts, it's potential energy is $q(100 \ Volts) = -e(100 \ Volts) = -1.60217662 \times 10^{-17} \  joules$. Negative plate is at $0$, positive is at some negative number, thus the electron moves in the direction of negative potential energy, the positive plate.
A: Potential is defined as
$V(r) = \int_{ref}^{r} \vec{E} \cdot \vec{dr}$
By convention, ref is set to $\infty$, in this reference point, ofcourse the charge has potential energy.
By setting the value of potential to be zero at the location of the plate, you change "ref" to be something other than infinity.
Does this actually matter?
No. The value of potential doesn't matter, since for all "ref", the change in potential is the same.
The value of ref just adds a constant onto the standard definition (when ref is set to $\infty$), which just cancels out when taking the difference in potential.
$(V_{0}(b)+c) - (V_{0}(a)+c)$
= $V_{0}(b) - V_{0}(a)$
This addition of a constant by changing the reference point of which potential is measured, also doesn't effect its relation to the electric field
Since,
$\vec{E} = - \nabla(V_{0}+c)$
$\vec{E} = - \nabla V_{0}$
