# Coulomb's Law and charge quantization

Coulomb's law states that the force between particles depends on their charge. But protons and electrons have equal but opposite charges. Shouldn't the formula simply have constants with the only changes required being the signs?

kuruman
Homework Helper
Gold Member
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In Coulomb's Law, the magnitude of the force is normally written as $$F=\frac{1}{4 \pi\epsilon_0}\frac{q_1q_2}{r^2}$$Considering that charges usually come in integral numbers of the electron charge magnitude ##e##, you could write $$F=\frac{1}{4 \pi\epsilon_0}\frac{N_1N_2e^2}{r^2}$$ for the magnitude, where ##N_1## and ##N_2## are positive integers. However, in E&M one usually models charge distributions as continuously smooth, i.e. charges can have fractional values of ##e##.

This idea is analogous to thinking of mass as being continuous without worrying about the "graininess" of atoms.

ZapperZ
Staff Emeritus
Coulomb's law states that the force between particles depends on their charge. But protons and electrons have equal but opposite charges. Shouldn't the formula simply have constants with the only changes required being the signs?

The flaw here is that you seem to think that "particles" can only be either a proton or an electron. There's nothing that says that those two are the ONLY entities that can use the label "particles". What if the particle is a composite particle, such as the nucleus of an atom? What is the force between a gold nucleus and an electron? What is the charge on an alpha particle? And wait till you hear that there are entities that can have fractional charges!

The word particles are not exclusively reserved only for protons and electrons.

Zz.

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Forgot about quarks, but I haven't studied them at all.

Ok I think follow. Thanks.

A follow on question... if you have 2 protons, then it's quite trivial to calculate. What if I have 2 clumps of protons. Are the charges simply additive? 3D interactions seem much more complex than individual particles.

jbriggs444
Homework Helper
A follow on question... if you have 2 protons, then it's quite trivial to calculate. What if I have 2 clumps of protons. Are the charges simply additive?

If you have a collection of charges, it becomes important to assign a position to the associated net charge. As long as the clumps are small compared to the distance between the clumps, you can wave your hands and treat the net charge as if it is located in the center of the clump. If the clumps are large compared to the distance between and if the charges are not distributed in a spherically symmetric manner then things get more complex.

Atomic_Sheep
jtbell
Mentor
Coulomb's law applies to point-charges. If the clumps are small, and the distance between them is much larger than their size, then it is a very good approximation to consider the clumps as point-charges.

Otherwise, you can proceed in two ways. If the charges in the clumps are themselves point charges, you can apply Coulomb's law to each pair of charges (one in one clump, one in the other), and add the results. If the charges are in a continuous distribution, you have to use integral calculus to perform the addition.

In three dimensions, you have to use vectors, because force is a vector quantity (magnitude and direction). $$\vec F = k \frac {q_1 q_2} {r_{12}^2}\hat r_{12}$$ where r12 is the distance between q1 and q2 and ##\hat r## is unit vector that points in the direction from q1 to q2.

Atomic_Sheep and jbriggs444
Thanks.