# Is negative potential energy actually meaningful?

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1. Oct 23, 2015

### kostoglotov

Disclaimer: not sure if this is actually the most appropriate forum, but it is a question inspired by my work on a homework question...I'm happy for a moderator to move this to a different forum. (But General Physics states:
NO HOMEWORK/COURSEWORK QUESTIONS HERE)

1. The problem statement, all variables and given/known data

Ok firstly, the question is, why is the potential energy of a system of two opposite charges negative?

2. Relevant equations

3. The attempt at a solution

FIRST: my answer is: with charges at infinity PE is zero, and the charges will move towards each other on their own, gaining KE, which must correspond to a decrease in PE, but it was zero, so it must go negative, so any system of two opposite charges with charges not at infinity must have negative PE...is this correct? and

SECOND:

IS Negative Potential Energy PHYSICALLY MEANINGFUL? Or is it just something we need to conceptualize or cope with in order to accommodate our mathematical models? Is this like the square root of negative one...in that, we just suppose that it can exist and then see what happens when we do?

edit: after all, if two opposite charges were at rest stuck to each other, and you pulled them apart, to have some distance r between them, then haven't you increased the potential energy of the system? Why wouldn't it make sense to think of the system of two opposite charges stuck together at rest as having zero potential energy?

Last edited: Oct 23, 2015
2. Oct 23, 2015

### andrewkirk

yes negative PE is meaningful. That's because whenever PE is used, it is only changes in PE that affect the physical outcomes. So we can add or subtract any constant amount we like to all potential energy measurements and it won't change the physical system.

Negative potential energies are often used because they provide a very convenient way of representing energy in a field that has an inverse square law.

3. Oct 23, 2015

### Mister T

True, but realize that it was an arbitrary choice to make it zero there. Just like in simpler cases where PE=mgy we can choose PE to be zero by choosing y = 0 at any height.

The point is this. If the potential energy happens to be zero, is it possible to lower it any further?

Of course it is, and that's when the potential energy becomes negative.

4. Oct 23, 2015

### SteamKing

Staff Emeritus
Potential energy is calculated with respect to some arbitrary datum. Negative PE just tells you on which side of this datum you happen to be.

5. Oct 24, 2015

### kostoglotov

So we could choose PE = 0 at the point where two unlike charges were at rest and touching each other.

6. Oct 24, 2015

### Mister T

No, not that point. But any other, yes.

The reason you cannot define the potential energy there is because you cannot define the force there. The magnitude of the Coulomb force is $k \frac{q_1q_2}{r^2}$, so it's value is undefined when $r=0$.

7. Oct 24, 2015

### kostoglotov

Hmm...sure, but in reality r will never equal 0...the r will simple be a function of the two radii of the charges, right? Of course, I'm not thinking about them as quantumn particles, etc. I'm sure it's not that simple :P

8. Oct 24, 2015

### Mister T

Well, we usually model them as point charges. Physically, that model is equivalent to assuming that $r$ is much larger than the sizes of the objects. Mathematically, we cannot define the potential energy to be zero when $r=0$.

In the usual scheme, described as setting the potential energy to zero "at infinity", we are actually saying that the limit of the potential energy, as $r \rightarrow \infty$, equals zero.

9. Oct 24, 2015

### kostoglotov

Can you help me understand why the voltages at all three points in this picture below are equal (and I think equal to zero): sources charges are equal magnitude

I can understand it from the algebraic approach, applying the formula for voltage $V = \frac{U}{q}$ and $U = K\frac{q_1 q_2}{r}$, but I don't get it conceptually.

Does it mean that if we started at any point at infinity and moved a charge to any of those three points (or presumably anywhere on the equipotential line formed by them), that the overall work done would be zero?

10. Oct 24, 2015

### Mister T

To see that they're all at the same potential note that the work done to move from one to another is zero. Precisely because they are all on the same equipotential line.

I wonder, though, if your conceptual difficulty doesn't lie with an improper application of $U=k \frac{q_1q_2}{r}$. Note that letting $q_1$ equal the charge on the red object and $q_2$ equal the charge on the blue object tells you the work done to assemble those two objects and has nothing to do with the points 1, 2, and 3.

11. Oct 24, 2015

### kostoglotov

No, I was applying that formula with $q_1$ as one of the source charges and $q_2$ as any possible charge at a point 1,2, or 3, and then I would find the V and add it to the V found by finding the V from the other source charge. Doing that gives zero at all points.

$U = K q \left(\frac{q_+}{r}-\frac{q_-}{r}\right)$

12. Oct 24, 2015

### Mister T

You mean $U = K q \left(\frac{q_+}{r}+\frac{q_-}{r}\right)$.

And you said $V=\frac{U}{q}$.

So,

$V=K \left(\frac{q_+}{r}+\frac{q_-}{r}\right)$.

Thus, for any point 1, 2, or 3, the potential due to q+ is equal but opposite to the potential due to q-.

13. Oct 25, 2015

### kostoglotov

But wouldn't you be ultimately subtracting the potentials from the source charges, since they are opposite charge?

14. Oct 25, 2015

### Mister T

Not the way you had it written.

Note that q-<0.

15. Oct 25, 2015

### kostoglotov

Ah well that was my intention, I was treating the q's as just magnitudes.

16. Oct 25, 2015

### Mister T

While we often do that in the formula for the Coulomb force, we almost never do it that way in the formula for the potential energy.

17. Oct 25, 2015

### Mister T

I can imagine a student's confusion if I wrote it that way. I'd be saying

"When I wrote q- (pronouncing it q minus) I meant that q was positive."