The Unapologetic Mathematician

Coulomb’s Law

I want to start in on a new topic, but it might be a bit of a surprise. I haven’t really talked about any applications much at all. Still, physics is a huge area of application for mathematics, and a lot of mathematics wouldn’t have been discovered without the physical motivation.

But why didn’t I talk about classical Newtonian mechanics when discussing calculus? As it happens, the application of calculus to Newtonian mechanics is pretty straightforward and boring; the first-pass coverage is pretty much all there is. Electromagnatism, however, is another story. The first-pass treatment is basically all about vector calculus, and that’s great; we’ll go over that a bit, which may be review for some people. But there’s a much deeper story to even classical electromagnetism that uses all this stuff I’ve been saying about differential geometry lately. But for now everything will take place in regular three-dimensional space.

Anyway, we start with Coulomb’s law. This is something that can be experimentally determined, but we’ll take it as an assertion — another axiom — and build from there. When we have two charged particles, they exert a force on each other. The magnitude of the force is proportional to the magnitude of the charge on each particle, and inversely proportional to the square of the distance between them. The direction of the force exerted on the first particle by the second is in the direction of the vector pointing from the second to the first if their charges have the same sign, and in the opposite direction if they have different signs. That is, charges of the same sign push each other apart, while charges of the opposite sign pull each other together.

Let’s write this out in a formula: if the charge on the two particles are $q$ and $q'$ — measured as a positive or negative multiple of some unit — and if the displacment vector from the second particle to the first is $r$, then we have the following formula for the magnitude of the force exerted on the first particle by the second:

$\displaystyle\lvert F\rvert=\frac{1}{4\pi\epsilon_0}\frac{\lvert q\rvert\lvert q'\rvert}{\lvert r\rvert^2}$

since the distance between the particles is given by the length of $r$. To get the direction, we use the unit vector $\hat{r}=\frac{r}{\lvert r\rvert}$ that points from the second particle to the first. It turns out that we also just need to drop the absolute value signs on the charges:

$\displaystyle F=\frac{1}{4\pi\epsilon_0}\frac{qq'}{\lvert r\rvert^2}\hat{r}=\frac{1}{4\pi\epsilon_0}\frac{qq'}{\lvert r\rvert^3}r$

Now, I haven’t explained why the constant of proportionality is written in the weird form $\frac{1}{4\pi\epsilon_0}$, and I’m not going to quite yet. I’ll just say that that’s all this is: a constant that gets the scaling right, not to mention the units. On the right, we’ve got (other than the constant) units of charge squared over area, while on the left we’ve got force, which is mass times distance over time squared. The “electric constant” $\epsilon_0$, thus, must carry units of time squared times charge squared over mass times volume.

In the common SI (metric) system we measure charge in coulombs — after Coulomb’s law — with symbol $\mathrm{C}$ and we have a convenience unit called the “farad” with symbol $\mathrm{F}$, which is given by

$\displaystyle\mathrm{F}=\frac{\mathrm{s}^2\cdot\mathrm{C}^2}{\mathrm{m}^2\cdot\mathrm{kg}}$

Using these units, we can write the electric constant with units of farads per meter. Incidentally, it has the measured value of approximately $\epsilon_0=8.85418782\times10^{-12}\frac{\mathrm{F}}{\mathrm{m}}$, but the exact value will be largely irrelevant to us.

January 3, 2012 -

1. “That is, charges of the same sign push each other apart, while charges of the same sign pull each other together.” The second ‘same sign’ should probably be ‘opposite sign’. =)

Comment by djdead | January 3, 2012 | Reply

2. Thanks; editing error.

Comment by John Armstrong | January 3, 2012 | Reply

3. Congratulations for choosing this topic! I have become recently interested in rigorous treatments of physics using mathematical arguments (and electromagnetism is something I know next to nothing about, so it would be refreshing to read about it – but classical mechanics is not that straightforward in my opinion, there are many deep results in there, see for example Noether’s theorem). However, you should also explain for the “outsiders” what physical concepts mean (as you use terms like “force” or “particle” without defining them first, even informally), as you did in your first blog posts with rational numbers, for example.

I also like that you don’t use the usual physics convention in which $\vec{r}$ is a vector and $r$ its magnitude, but rather the mathematical one where $r$ is the vector and $\lvert r \rvert$. I think we don’t need a special sign to remind us that a variable denotes a vector, similarly to the fact that we don’t need one for complex numbers.

Comment by Andrei | January 3, 2012 | Reply

4. Small correction: in the second paragraph there should be “and $\lvert r \rvert$ the magnitude”.

Comment by Andrei | January 3, 2012 | Reply

5. Ah, you mean in parallel with “if the charges are $q$ and $q'$?

Comment by John Armstrong | January 3, 2012 | Reply

• No, I corrected my own comment.

Comment by Andrei | January 3, 2012 | Reply

Comment by Andrew | January 3, 2012 | Reply

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