The equation: $U_E = Vq$ or $(Eq)d$.

This works the same way as the equation for gravitational potential energy: $mgh$

But to me, for charges of different signs, the potential energy also varies.

Because $d$ is measured from lowest point, usually negative plate, in the case of parallel plate for example, the direction that a particle moves in affect the distance it travel, and thus the potential energy it has.

clarification For example, a negatively charge particle, starting from 2v, would move upward towards the positive plate, which is higher potential. A positive particle would move downward. Since potential kinda is distance. The two particles would have different potential energy. But this is not shown in the equation.

See here, if the charged particle is placed at exactly the middle, sign will affect direction and thus affect the potential energy. enter image description here

Same for point charge: enter image description here

Consider $E = k\frac{q}{r^2}$. Because "r" is the distance between the center of the particle to where the test point is located and because in this case the particle is positive, the equation would really apply if the test point placed near this particle is also positive, meaning if released, the test point would go away from it. Then the distance is not "r".


  • 1
    $\begingroup$ sorry, I am not clear exactly what you are asking - can you be more specific about the concept you find counter-intuitive? $\endgroup$
    – tom
    Jul 26, 2015 at 21:35
  • $\begingroup$ Keep in mind as well ,the equation V=mgh for gravitational potential energy is an approximation for small h relative to the radius of the gravitating body (e.g. the Earth) $\endgroup$
    – paisanco
    Jul 26, 2015 at 22:00
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    $\begingroup$ What do you mean that the charge isn't in the potential? That's exactly what $q$ means. If $q>0$ then $Eqd$ has one sign, and if $q<0$ then it has the other sign. $\endgroup$
    – DanielSank
    Jul 26, 2015 at 22:27
  • $\begingroup$ Your question is unclear. Are you trying examine the behavior of electrical potential energy changes and gravitational potential energy changes on the same particle simultaneously as it changes position, or separately? The two potential energies (and potentials) arise from different forces and different properties (charge vs mass). $\endgroup$
    – Bill N
    Jul 26, 2015 at 22:38
  • $\begingroup$ Except that the particle is placed at exactly the middle between the two plates, because they travel in different direction, the distance travelled will differ. $\endgroup$ Jul 26, 2015 at 22:49

1 Answer 1


A particle subjected to only a conservative force field, without other restrictions, will move in a direction which reduces the potential energy of the system. The particle itself does not have potential energy; the particle-field system has potential energy (some may say the field-field system, but let's not nit-pick that yet).

So what's important is not the value of the potential energy of the system, but how the potential energy changes. Based on your formula we can write $$\Delta U_E = q\Delta V.$$ That means that a negative charge will move toward higher $V$ under the influence of the external $E$ field. A positive charge will move toward a lower $V$. The actual values of $V_{initial}$ and $V_{final}$ are irrelevant as long as $$V_{final}-V_{initial}=\Delta V$$ arithmetically has the correct sign.

An electron and a proton placed in the same field (separately, so they don't influence each other) will move in opposite directions. They are opposite sign and their unrestricted movements must reduce the system potential energy.

  • $\begingroup$ But in the case of a point charge, if a charge placed around it moves away from it say from 2V mark, you don't know exactly where it will stops, cand so can't calculate delta V. What do you do? $\endgroup$ Jul 27, 2015 at 14:31
  • $\begingroup$ The $V_{final}$ is potential at any particular point you want to choose. It's not the location where it "stops." It's really a 2nd location of your (or the problem's) choosing. And two charges moving away from each other with no other influence will never stop. $\endgroup$
    – Bill N
    Jul 27, 2015 at 14:38
  • $\begingroup$ If the speed of the particle is changing, when do we know the work is done ON or BY the particle? $\endgroup$ Jul 27, 2015 at 14:41
  • $\begingroup$ The work done is continually changing in that case. You have to pick a point that of interest to you. $\endgroup$
    – Bill N
    Jul 27, 2015 at 14:47
  • $\begingroup$ So when do we called that work done ON or BY the charge? $\endgroup$ Jul 27, 2015 at 14:48

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