This is how I understand Coulomb's Law's derivation, please let me know if it's correct.

Charles Augustin de Coulomb and some other scientists 'experimentally' deduced that there are three factors that affect the electrostatic force between two stationary charged particles with equal distribution of electric charge, those three factors are : the magnitude of charges (denoted by $q_1$ and $q_2$), the distance between the particles (denoted by $r$) and the medium, more specifically, it's permittivity (denoted by $\varepsilon _m$).
Again, experimentally, they deduced that : $$F_e \propto q_1q_2$$ $$F_e \propto \dfrac{1}{r^2}$$ $$F_e \propto \dfrac{1}{\varepsilon_m}$$

Combining these proportionalities, they deduced that : $$F_e \propto \dfrac{q_1q_2}{\varepsilon_m.r^2}$$ Now, that can be expressed as an equation with a constant of proportionality, let's call that constant $k$. $$F_e = k\dfrac{q_1q_2}{\varepsilon_m.r^2} \text{ and here }k = \dfrac{1}{4\pi} \text{ (in SI units) } \implies F_e = \dfrac{q_1q_2}{4\pi \varepsilon_m r^2}$$ Now, in this, $\dfrac{1}{4\pi \varepsilon_m}$ can be separated as $K_e$, Coulomb's constant (I read somewhere that $\dfrac{1}{4\pi\varepsilon_m}$ is chosen as $K_e$ because some people are 'comfortable' with it, but I don't think that's the case).

So, finally : $$F_e = \dfrac{q_1q_2}{4\pi\varepsilon_m r^2} = K_e \dfrac{q_1q_2}{r^2}$$

Is this derivation correct? I have read about another method of derivation as well, but the once that I mentioned above makes more sense to me. That 'another' method is : $$F_e \propto q_1q_2$$ $$F_e \propto \dfrac{1}{r^2}$$ $$\therefore F_e \propto \dfrac{q_1q_2}{r^2} \implies F_e = K_e \dfrac{q_1q_2}{r^2}$$ $$\text {Here, } K_e = \dfrac{1}{4\pi\varepsilon_m}$$

Let me know which one is correct, thanks!


1 Answer 1


Both are equally valid. The thing to notice is that the proportionality constant $K_{e}$ (as you have defined), is a medium specific constant. In your first derivation, you have factored this medium specific constant out and absorbed it in the beginning of your derivation, while in your second derivation you have added it in the end.

There is no correctness of one derivation over the other, as both are essentially the same. Pedagogically, the second one is preferred because it is simpler and doesn't involve too many things to consider. The permittivity of the medium is naturally defined and added once fundamental calculations are already done with vacuum.

Also, I have not encountered the statement that Coulomb experimentally deduced that the electrostatic force between two stationary charged particles depend on the permittivity. If I am correct, his experiments were more rudimentary and medium dependancy was a later addition. It'll be nice if you can cite some source.

  • $\begingroup$ I'm not so sure that Coulomb deduced that fact that permittivity affects the force, that's why I said 'Charles Augustin de Coulomb and other scientists' $\endgroup$ Jun 3, 2020 at 8:39
  • $\begingroup$ A question : If we derive the formula using the second method, then we get $K_e = \dfrac{1}{4\pi\varepsilon_m}$. Can you tell me how we arrive at this or share a link that would help me to understand this? I saw a video of Prof. Walter Lewin of MIT in which he said that $|pi$ is a part of $K_e$ just for historical reasons, is that true? $\endgroup$ Jun 3, 2020 at 8:46
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    $\begingroup$ The proportionality constant completely depends on the units you are working with. In SI units, you get $K_{e} = \frac{1}{4\pi\epsilon_m}$. And yes, $\pi$ is not necessary in this constant and can be absorbed into other equations by carefully redefining them. If you are trying to experimentally come up with all this, all you'll know is the force between two charged particles is weaker by some factor than in vacuum. This is what is important and this is what you try to incorporate. $\endgroup$ Jun 3, 2020 at 9:24
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    $\begingroup$ In other words, you'll just know that $K_e$ in some other medium, say water, is about 1/80 times that of $K_e$ in vacuum. All you want is your equation to express this observation. You can completely forego $1/4\pi$ and absorb everything into one constant. In the end, all your equations have to be self consistent. $\endgroup$ Jun 3, 2020 at 9:27
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    $\begingroup$ You can lookup Wikipedia's page on Coulomb's law here: en.wikipedia.org/wiki/Coulomb's_law, which has good information of how Coulomb's law is defined in different units. I hope this clarifies. $\endgroup$ Jun 3, 2020 at 9:28

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