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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".

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most venerable sir
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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
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    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
  • Think of equipotential lines. – most venerable sir Jul 26 '15 at 22:49
  • Note that you are comparing a "for use in limited circumstance (i.e. small motions near the surface of the some large mass) version of the gravitation rule to a general version of the rule for electrostatics. Of course the simplified case is simpler than the general case. – dmckee --- ex-moderator kitten Jul 27 '15 at 00:23
  • @dmckee, how is it simplified and still valid? – most venerable sir Jul 27 '15 at 02:08
  • http://physics.stackexchange.com/questions/6074/do-we-take-gravity-9-8-m-s%c2%b2-for-all-heights-when-solving-problems-why-or-why http://physics.stackexchange.com/questions/35878/is-acceleration-due-to-gravity-constant – dmckee --- ex-moderator kitten Jul 27 '15 at 02:16

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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.

Bill N
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  • 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? – most venerable sir Jul 27 '15 at 14:31
  • 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
  • If the speed of the particle is changing, when do we know the work is done ON or BY the particle? – most venerable sir Jul 27 '15 at 14:41
  • The work done is continually changing in that case. You have to pick a point that of interest to you. – Bill N Jul 27 '15 at 14:47
  • So when do we called that work done ON or BY the charge? – most venerable sir Jul 27 '15 at 14:48
  • Post that as a new question, after you do a search on Physics Stack Exchange to look and see if it has been asked and answered before. – Bill N Jul 27 '15 at 15:00
  • Is it always true that when net potential is zero, the net field strength, E, will also be zero? – most venerable sir Jul 27 '15 at 15:23
  • No, in fact, it's rarely true if there are any unbalanced charges present. E-field strength is the (negative of) slope of the potential. A zero E-field will be a local maximum or minimum of potential. And that's a new question, too! – Bill N Jul 27 '15 at 15:29
  • You guys always say the exact value of potential isn't as important as the potential difference. But in the case of resistors using up all potential/voltage supplied by the battery, say If the battery supply 3v of potential, then the final voltage must be 0, since there can't be any voltage left. (It can't go from 5v to 2v)? – most venerable sir Jul 27 '15 at 20:32
  • Voltage is difference of potential. And resistors don't "use up" potential. They change the potential. Resistors transform energy. – Bill N Jul 27 '15 at 22:11