Is it a coincidence that the electric potential energy of two point charges is equal to the force between them times the distance?

Question

Take a point charge $q_1$. Its potential is $V = q_1/4\pi\epsilon_0r$. Therefore the potential energy of a second point charge $q_2$, at distance $r$, is $PE = q_1q_2/4\pi\epsilon_0r$. This expression is curiously (?) equal to the electric force between the two $F = q_1q_2/4\pi\epsilon_0r^2$, multiplied by the distance $r$:

$$\frac{q_1}{4\pi\epsilon_0r} \times q_2 = \frac{q_1q_2}{4\pi\epsilon_0r^2} \times r$$

Is there a reason why multiplying the force between two charged objects & their distance yields the potential energy? If there is a reason, I'm not seeing it: the potential energy is given by the work done to take the two from infinitely far away to distance $r$ away, or $\int F dr$. But the force changes throughout the range, so I still don't see why we can take the final force and multiply that by $r$. $r$ is even in the "wrong direction" since it's the distance towards the other charge, not the distance from infinity.

I'm looking for an explanation at the intuitive level. The math clearly works out, I just don't see why it works out, unless it's a coincidence.

Edit: To illustrate the issue I don't understand, the potential energy is supposed to be the work done to take the two objects from infinitely far away to distance $r$ away. Therefore it seems like this is the appropriate way to calculate the potential energy:

$$PE = \int^{\infty}_r\frac{q_1q_2}{4\pi\epsilon_0r^2} dr$$

However this is equal to the expression

$$\frac{q_1q_2}{4\pi\epsilon_0r^2} \times r$$

Here we are taking the final force, multiplied by the final distance between the two charges. I can't see any reason why this force $\times$ distance should yield the work done. The charges don't move a distance $r$, and they certainly don't experience the force $\frac{q_1q_2}{4\pi\epsilon_0r^2}$ (where $r$ is the final distance) as they approach each other, since the distance between them is constantly changing.

I'm looking for an explanation why this force $\times$ distance still yields the potential energy. Like, it seems to me that to calculate the potential energy, you must use calculus, but the second expression doesn't seem to use calculus at all, and it still works. Why?

Answer 1

It is not a coincidence. They are related because voltage is defined with respect to energy

$$V=-\int_\mathcal C E\cdot dl$$

where $\mathcal C$ is an arbitrary path from some reference point (often a point infinitely far away)

Do note that voltage is not a force. It is a scalar quantity, rather than a vector quantity. We often think of voltage as how much "force" the electricity has, but that is just a convenient mental image, not a physically accurate choice of words.

Answer 2

It follows from the mathematical fact that $$ \frac{d}{dr}\frac{1}{r}=\frac{-1}{r^2}=-\frac{1}{r} / r.$$ That is, on the one hand $$ F(r)=\frac{q_1q_2}{r^2}=-\frac{d}{dr} \frac{q_1q_2}{r}=-\frac{d}{dr}U(r), $$ whereas on the other hand $$ F(r)=\frac{q_1q_2}{r^2}=\frac{U(r)}{r}. $$

In other words, there are two factors behind this:

• The electrostatic force is the gradient of the electrostatic potential energy, and conversely potential energy is an integral from force (by definition of a potential energy.)
• Electrostatic force behaves as $\frac{1}{r^2}$, which is a consequence of the Gauss law.

Related:
Why is Coulomb's law is an inverse square law?
Does Coulomb's Law, with Gauss's Law, imply the existence of only three spatial dimensions?

Answer 3

Using the magnitudes from the problem, which are two electric charges $q_1, q_2$ at a distance $r$, the only quantity with units of energy that you can write —up to a certain multiplicative constant— is precisely (check this yourself!) $$ \dfrac{q_1 q_2}{4\pi r}. $$ Of course, it's only natural that this is just $\propto Fr$, as you don't really have any more magnitudes relevant to the problem, so you can't write any other energy.

Now, about why we don't get a different result, proportional to this one —like $\frac{q_1 q_2}{8\pi r}$—, I think that you need to compute the integral that you can see in other answers, or even your question.

Answer 4

Although $r'\left(-\frac{dV}{dr}\right)_{r=r'}$ is indisputably a force multiplied by a distance, I don't think there's a physical (as opposed to a mathematical) reason why this should be equal to the work done per unit charge to bring a test charge from $r=\infty$ to $r=r'$.

You might like a more geometrical approach to the mathematics... Knowing that the slope of the $V=\frac Kr$ against $r$ graph at $r=r'$ is $-\frac K{r'^2}$, it's easy to show that the tangent to the $V=\frac Kr$ graph at $r=r'$ hits the $V$ axis at the point $\left(0, 2\frac K{r'})\right)$.

Therefore (easier to see with a sketch), looking at the portion of the tangent line between $r=0$ and $r=r'$, its slope is: $$\left(\frac{dV}{dr}\right)_{r=r'}=-\frac{2\frac K{r'}-\frac K{r'}}{r'}=-\frac{\frac K{r'}}{r'}=-\frac {V(r')}{r'}$$

Answer 5

If you forget about $q_1$, $q_2$, $\epsilon_0$, etc. and just look at the following integral equation:

$$\int_\infty^r 1/x^2 \,dx= -1/r$$

Integrating $\bf{any}$ inverse-square function $f$ from $\infty$ to $r$, as one would do to calculate the associated potential, is the equivalent of multiplying the function $f(r)$ by $-r$.

Is this a "coincidence" or can some further math intuition be applied here to unpeel another layer of understanding? Not sure.

Answer 6

The reason is that the derivative of $1/r$ is $-1/r^2$.