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?

  • $\begingroup$ "Set the plate to zero potential", the charge has 0 potential energy at this location. The value of potential energy doesn't matter, the change in potential between any 2 locations DOES matter. I can choose my reference of potential to be any value I want, to change the value of potential how I see fit. $\endgroup$ Commented Apr 5, 2022 at 0:38

2 Answers 2


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.

  • $\begingroup$ Oh, so the convention of the negative plate having zero electric potential does not always apply then. Since you said I can place the zero value wherever I want as long as change stays the same, then I could have the negative plate have 100 volts and the positive plate have zero. This would make a lot more sense because then you have zero potential energy at the positive plate, which is in agreement with the fact that the electron is at maximum speed at the positive plate and thus has maximum kinetic energy (no potential energy then) $\endgroup$ Commented Jun 28, 2018 at 7:52
  • $\begingroup$ One thing is that the negative plate should always have less electric potential then the positive plate. This is fine because remember that to get potential energy you multiply by the charge, which for the electron is negative. The electric potential of the positive plate is $>$ than that of the negative plate, but the potential energy of the positive plate is $<$ than that of the negative plate, due to the negative charge. So the negative would be at 100 volts, postive at 200 Volts. $\endgroup$
    – Mike Flynn
    Commented Jun 28, 2018 at 16:10

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


$\vec{E} = - \nabla(V_{0}+c)$

$\vec{E} = - \nabla V_{0}$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.