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Is the formula for electric potential energy ($U = kqQ/r$) measured in absolute value? In other words, as the magnitude of $U$ increases, does electric potential energy increases too? For example, if $q<0$ and $Q>0$, as either one of them increase in magnitude, such as when $q$ becomes more negative, the magnitude of $U$ increases but it gets more negative. So does that mean electric potential energy increases as $q$ or $Q$ increases?

In another sense, if $q<0$ and $Q>0$, as $r$ increases: the magnitude of $U$ decreases (becomes less negative). So does electric potential energy decrease when r increases?

What approach do we take when looking at electric potential? The number line approach or the absolute value approach?

($U$ is potential energy, $k$ is a constant, $q$ and $Q$ are electric charge, and $r$ is the distance between $q$ and $Q$)

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  • $\begingroup$ Please define your variables. What are U, k, q, Q, and r? I can guess about them but I'd be more likely to give an answer that helps you if you are explicit about what they are. $\endgroup$
    – The Photon
    Sep 16, 2017 at 21:38
  • $\begingroup$ U = Potential Energy, k is a constant, q = electric charge, Q = electric charge, r = distance between q and Q. $\endgroup$
    – A.AK
    Sep 16, 2017 at 21:39
  • $\begingroup$ Also, be sure to understand the difference between electrostatic potential energy, and electrostatic potential. In paragraph 1 you talk about potential energy, but in paragraph 2 you talk about potential. These are two different things. $\endgroup$
    – The Photon
    Sep 16, 2017 at 21:40
  • $\begingroup$ I meant to say electric potential energy in that 2nd paragraph. Sorry $\endgroup$
    – A.AK
    Sep 16, 2017 at 21:40
  • $\begingroup$ Also, please include all information needed to answer your question in the question (you can edit it at any time). New readers shouldn't have to read comments to understand the question. $\endgroup$
    – The Photon
    Sep 16, 2017 at 21:41

2 Answers 2

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In another sense, if q<0 and Q>0, as r increases: the magnitude of U decreases (becomes less negative). So does electric potential energy decrease when r increases?

A value becoming less negative means that value is increasing.

This makes sense, since to separate two opposite charges you have to apply a force in the same direction as you're trying to move the charges. This means you're doing work on the system, and increasing its potential energy.

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  • $\begingroup$ Ok. So using the number line approach, if we increase q (negatively), U becomes more negative. That means potential energy decreases. Which doesn't make sense. $\endgroup$
    – A.AK
    Sep 16, 2017 at 21:53
  • $\begingroup$ Sure it does. Because if you have a larger charge $q$, it takes more work to push it away to infinity (where the potential energy would be 0). $\endgroup$
    – The Photon
    Sep 16, 2017 at 21:54
  • $\begingroup$ Or thought of another way, if you start the charge $q$ at infinity and let it push you towards $Q$ until reaching distance $r$ it does more work on you if $q$ has a larger magnitude. It has given up more potential energy to do that work on you. $\endgroup$
    – The Photon
    Sep 16, 2017 at 21:55
  • $\begingroup$ if you have to do more work to push something away to a certain distance, doesn't that mean the potential energy of that particle is greater than the potential energy of another particle that takes less work to push away to that same distance? In this case pushing q to a distance of r, compared to pushing a larger q to a distance of r. Wouldn't that mean that the larger q would have a greater potential energy (less negative). $\endgroup$
    – A.AK
    Sep 16, 2017 at 21:59
  • $\begingroup$ No, if you push it the same direction it's moving, that means you're doing work on it. You're adding to its potential energy. And for your equation to hold, we know we're defining the potential energy to be 0 at a distance of infinity. So it must have started at a lower (more negative) p.e. if you did more work on it to get it to zero. $\endgroup$
    – The Photon
    Sep 16, 2017 at 22:01
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The formula that you have quoted $U=k\frac{qQ}{r}$is the potential energy of two charges $Q$ and $q$ where there separation is $r$ and the potential energy is zero when the separation of the charges is infinite.

The sign that you assign to the charges is important and their product of the charges with signs included will produce a potential energy on a number line.

So if the two charges have the same sign an external force has to do work to bring the two charges from an infinite separation to a separation of $r$.
This means that the system of two charges has more potential energy (it is positive, greater than zero) when their separation is $r$ compared with their separation being infinite.

If the charges have opposite signs an external force needs to do work in separating the charges from $r$ to infinity thus the potential energy of the system is greater when separation of the charges is infinite than when their separation is $r$.
Since the potential energy of the two charges is zero at infinite separation the potential energy at separation $r$ must be negative.

A potential energy of $-100\,\rm J$ is lower than a potential energy of $+100\,\rm J$ and the difference is $200\,\rm J$.

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  • $\begingroup$ Why is electrical potential decreasing (becoming more negative) when either the charge q is becoming more negative or, Q is becoming more positive. The force of attraction between the two particles increases and thus requires more work to move to a certain distance r. So if we increase Q or q in magnitude, why does U become more negative (potential energy decreases. I don't understand it conceptually. $\endgroup$
    – A.AK
    Sep 16, 2017 at 22:10
  • $\begingroup$ @A.AK Suppose you have two charges $+Q$ and $-q$. They have a negative potential energy. If you repeat the process of bringing the charges from infinity to a separation $r$ when either the positive or negative charges are increased the system will have a lower potential energy ie more work has to be done separating the charges to infinity. $\endgroup$
    – Farcher
    Sep 16, 2017 at 22:21

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