This was supposed to be a long question but something went wrong and everything I typed was lost. Here goes.

  1. Why is $k = \dfrac{1}{4\pi\epsilon_0}$ in Coulomb's law?

  2. Is this an experimental fact?

  3. If not, what is the significance of this definition?

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    $\begingroup$ It's a feature of the choice of units (i.e. in other systems of units the constant can be 1 or $1/4\pi$). There are a number of existing questions that relate to this matter, and it may be a duplicate. Looking for a link... $\endgroup$ May 20, 2012 at 18:51
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    $\begingroup$ Here we go: physics.stackexchange.com/q/24505 , physics.stackexchange.com/q/1673, and maybe others. Let me know if those fail to answer your question. $\endgroup$ May 20, 2012 at 18:57
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    $\begingroup$ Tell that to the Gaussian units people. You can fold those values into the charge if you want. I don't, but it made sense to some people. $\endgroup$ May 21, 2012 at 13:57
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    $\begingroup$ @Ron The gravitational constant $G$ involves as much a choice of units as does Coulomb's law (in this case, setting gravitational mass strictly equal - rather than simply proportional - to inertial mass). $G$ can also be written as $1/4\pi\gamma_0$, and if you could ever make a gravitational capacitor then $\gamma_0$ would be the "permittivity" of the vacuum. Since $k$ and $\epsilon_0$ are (so rigidly) proportional, they share all their physical meaning. $\endgroup$ May 21, 2012 at 23:42
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    $\begingroup$ possible duplicate of Why is there a factor of $4\pi$ in certain force equations? $\endgroup$ Sep 18, 2013 at 9:12

3 Answers 3


Defining the symbol $k$ in Coulomb's law, $$F=k\frac{q_1q_2}{r^2},$$ to be $k=1/4\pi\epsilon_0$, is perfectly allowed when one understands it simply as a definition of $\epsilon_0$. The motivation for this definition is that when you work out the forces between two oppositely charged plates of area $A$ and charge $Q$ a distance $d$ apart, they come out as $F=\frac{2\pi kQ^2}{d}=\frac{Q^2}{2\epsilon_0 d}$, where the factor of $4\pi$ comes from judicious application of Gauss's law.

When you develop this further into a theory of capacitance, you find that it implies the voltage between the plates is $V=Q/C$, where $C=\epsilon_0 A/d$. Further, if you want to insert a dielectric in between the plates (as you often do), then the capacitance changes to $$C=\epsilon A/d$$ where $\epsilon$ is known as the dielectric's electric permittivity. $\epsilon_0$ is then naturally understood as "the permittivity of free space" (which of course simply defines what we mean by permittivity).

The question is then, of course, why is this "derived" unit, $\epsilon_0$, treated as more "fundamental" than the original $k$? The answer is that it isn't since they are equivalent, but the permittivity of free space is far easier to measure (and certainly was so during the late 19th and early 20th centuries when electrical research was very much geared towards circuit-based technologies), so that it came out the winner, and why have two symbols for equivalent quantities?


The unit of the second is defined is the time duration of a certain number of periods of radiation emitted from a particular type of electron transition between energy levels in an isotype of Cesium (see here).

It is an assumption that light travels at a constant speed $c$ independent of one's reference frame, so now that we have fixed a unit of time, we may define a unit of length: the meter is the distance light travels in $1/299792548\, \mathrm{s}$.

We also define the SI unit of current (the Ampere) so that the permeability of free space takes on a desired value in SI units ($4\pi \times 10^{-7}$).

We may then define $$ \varepsilon _0=\frac{1}{\mu _0c^2} $$ as well as $$ k=\frac{1}{4\pi \varepsilon _0}. $$

Now, keep in mind, you do not have to fix a system of units to do this (as I did before). As the above are definitions, they will hold in any system of units. However, to see that these definitions do not wind up being circular, it helps to see that we can define $\mu _0$ and $c$ in terms of purely physical phenomena. In other words, for the above definitions to even make sense, we had to know that we could define $c$ and $\mu _0$ independent of $\varepsilon _0$ and $k$ first. The above definition of SI units helps you see that this can be done.

  • $\begingroup$ This is all changing with the new system of SI. While $c$ is fixed, $\mu_0$ and $\epsilon_0$ are not. $\endgroup$
    – ProfRob
    Nov 10, 2018 at 17:32

If the question is why the "$4\pi$" in the Coulomb constant (k=$\frac{1}{4\pi\epsilon_{0}}$), then an equally valid question could be why the "4$\pi$" in the magnetic permeability of vacuum, $\mu_{0}=4\pi\times10^{-7}H/m$ ?

Perhaps a clue can be found in the Maxwell's equation for the speed of electromagnetic wave (light) in a vacuum, $c=\frac{1}{\sqrt{\epsilon_{0}\mu_{0}}}$.

Of course, Maxwell derived this relationship much later than Coulomb.

Maxwell relates the electric permitivity to magnetic permeability in the vacuum, $\mu_{0}=\frac{1}{\epsilon_{0}c^{2}}$ which is given a value of $\mu_{0}=4\pi\times10^{-7}H/m$ in SI units.

The 'reason' for the "$4\pi$" appearing here and in Coulomb's constant (believe it or not) so that Maxwell's equations can be written without any $4\pi$' factors!

In order to understand this, consider how electrostatic phenomena are expressed in Coulombs law as "field intensity at a distance squared", compared to (the equivalent) Gauss' law, which describes the "flux through a closed surface enclosing the charge".

The total flux is the flux density multiplied by the surface area, which for a sphere of radius $r$ is given by $S=4\pi r^{2}$, so the ratio $S/r^{2}$ = $4\pi$ is simply the result of geometry of space and spherical symmetry.

The SI system of units (unlike the Gauss units) is said to be 'rationalized' because it allows the expression of Maxwell's equations without the $4\pi$ factors. To do this, the $4\pi$ factor has simply been "built into" the (SI unit) definition of the universal constant for permeability of the vacuum, $\mu_{0}=4\pi\times10^{-7}H/m$, from which we can express Coulomb's constant as k=$\frac{1}{4\pi\epsilon_{0}}$.


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