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We know that in Coulomb's law, the constant $k_e$ is $\frac{1}{4\pi \epsilon_0}$. Where does $4\pi$ come from? Why is it related to the force between two charges?

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    $\begingroup$ The surface area of a unit sphere is $4\pi$. Do you know about Gauss's law? $\endgroup$
    – J.G.
    Commented Apr 16 at 8:58
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    $\begingroup$ Related: physics.stackexchange.com/q/605644/2451 and links therein. $\endgroup$
    – Qmechanic
    Commented Apr 16 at 9:31
  • $\begingroup$ Possible duplicate. $\endgroup$
    – joseph h
    Commented Apr 16 at 10:23
  • $\begingroup$ A surface of constant electric field strength that is generated by a point charge is spherically shaped. The surface area of that sphere is $4\pi r^2$. $\endgroup$ Commented Apr 16 at 14:30

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The point is not so much about the factor of $4\pi$ being in the force, but rather that there must be a factor of $4\pi$ between the force (i.e. "Coulomb's law") and divergence (i.e. "Gauss's law") for the field since by the divergence theorem, we have the following identity for inverse-square fields: $$\nabla\cdot\frac{\hat{\mathbf{r}}}{r^2} = 4\pi\delta(\mathbf{r}).$$

It is not necessary for the $4\pi$ to be in the "Coulomb's law" for the field. In Newtonian gravity, it appears in the divergence instead since $\nabla\cdot\mathbf{g} = -4\pi G\rho$. In Gaussian units, it is also moved to the divergence, giving $\nabla\cdot\mathbf{E} = 4\pi\rho$. I believe that the placement is a matter of convention, much like the factor of $2\pi$ between the Fourier transform and the inverse Fourier transform.

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It's a choice. The SI is a rationalized unit system, which means that in electromagnetic formulae, factors of $4\pi$ appear in radial formulae, while they don't appear in rectangular formulae. It's not a necessary choice: in Gaussian units, $k_e$ is $1$, which simplifies spherical formulae, but causes $4\pi$ to appear in rectangular formulae.

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Gauss law: Electric field inside a closed surface $S$ is proportional to the surface charge.

$$E = \frac{Q/S}{\varepsilon}$$

In the case of an infinite uniformly charged plane you will get

$$E=\frac{Q/2S}{\varepsilon}$$ where 2 comes from two sides of the plane.

In the case of a point charge, the field at a distance $R$ from the charge is

$$E = \frac{Q/(4\pi R^2)}{\varepsilon}$$ because sphere with radius $R$ has surface $4\pi R^2$.

In this case you can reformat this equation to get electric field of a point charge at distance $R$ and see where Coulomb constant comes from: $$E = \frac{1}{4\pi\varepsilon}\frac{Q}{R^2}=k_e\frac{Q}{R^2}$$

And finally, the force: $\overrightarrow F=q\overrightarrow E=k_e\frac{qQ}{R^2}\hat R$ with outward (repelling) direction for positive $q$ and $Q$.

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The force between two charges is dependent on the distance between them. The specific positions does not really matter but only the distance (in this simple two-charge scenario).

In other words, if you trace a charge on a sphere around another charge, the force will be (in the case of opposite sign charges) if equal magnitude in all cases, and directed towards the central charge. There is spherical symmetry in other words. Therefore it is modeled that the force between them is the product of the two charges divided by the area of a sphere with a radius of the distance between the charges, and there the pi comes from.

$F = \frac{Q}{4\pi\epsilon_0r^2} = q_1\frac{q_2}{4\pi\epsilon_0r^2} = q_2\frac{q_1}{4\pi\epsilon_0r^2} = \frac{\text{product of charges}}{\text{area of sphere}}$

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If we consider the Gauss’s equation; Electric flux through a body is equal to the charge passing divided by its permittivity in free space. This gives us the equation: Flux = Charge/Permittivity of free space

Flux can also be written as: $E.ds$

This gives us, $E.ds$ = Charge/Permittivity

We know that the electric field experienced will be the same for all charges in space at a distance r from a test charge.

So considering a positive charge which emits radially outward field line in uniformity, we assume the simplest shape for measuring flux about scattered points at uniform distance from a test charge: i.e. a sphere. (Keep in mind that we can use all sorts of other geometrical shapes but sphere is the simplest)

Hence taking cyclic integral of the above equation: $E \times 4 \pi r^2$ (Surface area of a sphere) = $q/\epsilon_0$ of free space

So we get electric field to be: $E = \dfrac{1}{4 \pi \epsilon_0} \dfrac{ q}{r^2}$

And hence multiplying by charge gives us the force between the two.

Also sorry I couldn’t find how to use the symbols for everything :(

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    $\begingroup$ This doesn't introduce anything new to the answers that already appeared. $\endgroup$
    – Radek D
    Commented Apr 17 at 9:07

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