# Why is the equation for electric potential energy so counter-intuitive?

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. Same for point charge: 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".

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• sorry, I am not clear exactly what you are asking - can you be more specific about the concept you find counter-intuitive? – tom Jul 26 '15 at 21:35
• 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) – paisanco Jul 26 '15 at 22:00
• 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. – DanielSank Jul 26 '15 at 22:27
• 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). – Bill N Jul 26 '15 at 22:38
• 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. – most venerable sir Jul 26 '15 at 22:49

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.
• 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. – Bill N Jul 27 '15 at 14:38