Potential energy of an electrostatic system in equilibrium

[Rob2024]

Homework Statement

Purcell and Morin Electromagnetism Problem 1.6

In view of the previous result, we might make the follow-
ing conjecture: “The total potential energy of any system of
charges in equilibrium is zero.” Prove that this conjecture is
indeed true. Hint: The goal is to show that zero work is required
to move the charges out to infinity. Since the electrostatic force
is conservative, you need only show that the work is zero for
one particular set of paths of the charges. And there is indeed
a particular set of paths that makes the result clear.

Relevant Equations

N/A

The solution given is as following:
Consider an arbitrary set of charges in equilibrium, and imagine moving them out to infinity by uniformly expanding the size of the configuration, so that all relative distances stay the same. For example, in part (b) we will simply expand the equilateral triangle until it becomes infinitely large. At a later time, let $f$ be the factor by which all distances have increased. Then because the electrostatic force is proportional to $1/r^2$, the forces between all pairs of charges have decreased by a factor $1/f^2$. So the net force on any charge is $1/f^2$ of what it was at the start. But it was zero at the start, so it is zero at any later time. Therefore, since the force on any charge is always zero, zero work is needed to bring it out to infinity. The initial potential energy of the system is thus zero, as desired. (You can quickly show with a counter example that the converse of our result is not true.)

Is there another way to prove the conjecture that does not use the above 'expanding to infinity' argument?

 

[kuruman]

Please explain what is going on in part (b). It looks like there is an equilateral triangle, presumably with charges at its corners. Is that all there is?

 

[Rob2024]

In part (b), there is an equilateral triangle with 4 charges. There is a big $Q$ at the center with distance $d$ to small $-q$ at each vertex.

 

[kuruman]

In part (b), there is an equilateral triangle with 4 charges. There is a big $Q$ at the center with distance $d$ to small $-q$ at each vertex.

OK, for a specific value of $Q$ in terms of $q$ you can have the net force on any one of the four charges be zero. This happens when $~Q=\frac{1}{\sqrt{3}}q.$ Note that this condition is independent of the size of the triangle. This means that the net force on each charge is zero for any size of the triangle including infinitely large.

Note that if I let the side of the triangle be $a$, the distance between $+Q$ and any of the charges $-q~$ is $~d=\frac{1}{\sqrt{3}}a.$ You don't need the infinity argument, just calculate the electrostatic potential energy of the configuration relative to infinity. It should be zero.

Please note that Purcell & Morin's use of the word "equilibrium" could be misleading because they do not clarify that the equilibrium is unstable. I am sure that Purcell and Morin are aware of Earnshaw's theorem that "a collection of point charges cannot be maintained in a stable stationary equilibrium configuration solely by the electrostatic interaction of the charges."

In my opinion, the authors should have put forth a conjecture that omits mentioning equilibrium. For example, “In a system of charges where the net electrostatic force on each charge is zero, the potential energy of that system is zero.”

 

[Rob2024]

That's correct, it was made clear the equilibrium is unstable from the context of the problem. The net force must be 0 if the charges are in equilibrium (no matter which kind). What bothers me is the infinity proof, it feels unsatisfactory. The triangular setup is only meant as an example, the conjecture is not limited to the triangular setup.

 

[haruspex]

The triangular setup is only meant as an example

… and it was only used as an example. The argument laid out in post #1 is quite general.

 

[Steve4Physics]

... let $f$ be the factor by which all distances have increased. Then because the electrostatic force is proportional to $1/r^2$, the forces between all pairs of charges have decreased by a factor $1/f^2$. So the net force on any charge is $1/f^2$ of what it was at the start. But it was zero at the start, so it is zero at any later time. Therefore, since the force on any charge is always zero, zero work is needed to bring it out to infinity. The initial potential energy of the system is thus zero, as desired.

It may be worth noting that (AFAICS) the same argument would be valid for any power law, $F \propto \frac 1{r^n}$.

 

[Rob2024]

Would it be possible to prove for an arbitrary collection of charges $q_i$ at $\vec r_i$ in equilibrium, the total potential energy is 0. For a concrete configuration, it's calculable, but what if the configuration is more abstract as proposed? Is there a way to show the total PE is 0 through direct calculation of $\sum_{i,j} \frac{k q_i q_j}{2 \Big|{\vec r_i - \vec r_j}\Big|}$?

 

[haruspex]

Would it be possible to prove for an arbitrary collection of charges $q_i$ at $\vec r_i$ in equilibrium, the total potential energy is 0.

Yes, by formalising the handwaving proof in post #1.

Is there a way to show the total PE is 0 through direct calculation of $\sum_{i,j} \frac{k q_i q_j}{2 \Big|{\vec r_i - \vec r_j}\Big|}$?

You could try writing the equations that represent the equilibrium as a start.