How do I visualize total potential energy of charges?

Question

In the following exercise from University Physics with Modern Physics (Young, Hugh D.; Freedman, Roger A..)

Two point charges are at fixed positions on the $x$-axis, $q_1 = -e$ at $x = 0$ and $q_2 = +e$ at $x = a$.

(a) Find the work that must be done by an external force to bring a third point charge $q_3 = +e$ from infinity to $x = 2a$.

(b) Find the total potential energy of the system of three charges.

the answer is

(b) $U=-\dfrac{e^2}{8 \pi \epsilon_{0} a}$

Our negative result in part (b) means that the system has lower potential energy than it would if the three charges were infinitely far apart. An external force would have to do negative work to bring the three charges from infinity to assemble this entire arrangement and would have to do positive work to move the three charges back to infinity.

I am finding it difficult to visualize negative work when the total potential energy of the system is negative. According to the textbook, it states that $U_a - U_b$ is the work that must be done by an external force to move the particle slowly from $b$ to $a$ against the electric force.

So I understand how "an external force would have to do negative work to bring the three charges from infinity to assemble this entire arrangement and would have to do positive work to move the three charges back to infinity" but I do not know how to visualize it.

I think that the displacement and force have to be acting opposite to each other for there to be negative work. I'm not too sure on how to apply this conception to a system of particles.

Answer 1

The order in which the charges are brought together does not matter so lets start with charge $+e$ at position $x=a$ and bring in charge $+e$ to position $x=2a$ from infinity.

No problem here I hope in that there is a repulsive electrostatic force between the two charges and external work needs to be done to bring the system of two positive charges together.
The potential energy of the system of two charges is $U_{\rm +e(a),+e(2a)} = +\dfrac{e^2}{4 \pi \epsilon_{0} a}$.
What this means is that the system is able to do work when the charges are free to separate.

Now bring in the third $-e$ charge from infinity to $x=0$ and the potential energy of the system of three charges is now $U_{\rm +e(a),+e(2a),-e(0)} = -\dfrac{e^2}{8 \pi \epsilon_{0} a}$.
What is the effect of bringing in the third charge?
Aa the third negative charge was brought in it was attracted to the other two positive charges and so could do positive work, eg the negative charge acted on by the attractive forces could be gaining kinetic energy.
In terms of the system of three charges where has the gain in kinetic energy of the third charge come from?
It has come from the decrease in the potential energy of the system of three charges and the decrease is so large that the potential energy of the system becomes negative.
If in some way the negative charge does external work such that it stops at $x=0$ then you have a system of three charges which requires external work to be done to make them separate to infinity where $U_{\rm +e(\infty),+e(\infty),-e(\infty)} =0$.

Lets look at the disassembly of the charges in energy terms.

$U_{\rm final}- U_{\rm initial} =U_{\rm +e(\infty),+e(\infty),-e(\infty)}-U_{\rm +e(a),+e(2a),-e(0)}= \text{external work done to separate charges}$

$0-\left(-\dfrac{e^2}{8 \pi \epsilon_{0} a} \right) = +\dfrac{e^2}{8 \pi \epsilon_{0} a} =\text{external work done to separate charges}$

which you interpret as the system of three charges doing negative work.