The electric field of a point charge $q$ is well known to be $$\mathbf E=\frac{q}{4\pi\epsilon_0 |\mathbf r|^3}\hat{\mathbf r}$$

This can be derived easily from integral form of Gauss’s law. Taking $V$ as a sphere of radius $r$ centered at the point charge, $$\int_{\partial V}\mathbf E\cdot d\mathbf A=\frac{q}{\epsilon_0}$$ $$4\pi r^2 E=\frac{q}{\epsilon_0}\implies E=\frac{q}{4\pi\epsilon_0 r^2}$$

However, I can’t see how this can be derived directly from the differential form of Maxwell’s equations. Maxwell’s equations only says the divergence and curl of electric field is zero everywhere away from the point charge, which isn’t quite useful at all.

How can the formula for electric field of a point charge derived directly from differential form of Maxwell’s equations?

(Note: it is of course possible to convert the differential form into integral form and the proceed as I’ve shown. However, that’s somewhat trivial and I do not consider that as a direct derivation.)

  • $\begingroup$ @AaronStevens My emphasis is on ‘differential form’ and ‘derived directly’. $\endgroup$
    – Szeto
    Oct 13, 2018 at 4:24
  • 2
    $\begingroup$ The charge distribution is not continuous and the differential form of Maxwell's equations would involve a kind of Dirac delta function for the charge density, since it is infinite at the origin and zero everywhere else. $\endgroup$
    – Tob Ernack
    Oct 13, 2018 at 4:31
  • $\begingroup$ To add to what @TobErnack it's saying, the fact that the divergence is not $0$ at the point charge should be useful. $\endgroup$ Oct 13, 2018 at 4:34
  • $\begingroup$ @TobErnack But how can this be useful for the derivation of the formula of electric field? $\endgroup$
    – Szeto
    Oct 13, 2018 at 4:52

2 Answers 2


When you have a point charge, the charge density is not really defined, as it is basically infinite at the origin, and zero everywhere else. You would need a distribution rather than a function to describe it. The derivation here is not completely rigorous as I am treating $\delta_{\mathbb{R}^3}(\vec{r})$ as a normal function when doing the integrations...

We can use the Dirac delta function over $\mathbb{R}^3$ to write $$\rho(\vec{r}) = q\delta_{\mathbb{R}^3}(\vec{r})$$

Then the Gauss equation gives $$\nabla \cdot \vec{E} = \frac{q}{\varepsilon_0}\delta_{\mathbb{R}^3}(\vec{r})$$

Since the problem is spherically symmetric, we compute the divergence in spherical coordinates and use $\vec{E} = E_r \hat{r}$ (where $E_r$ is the radial component of $\vec{E}$ and $\hat{r}$ is a unit radial vector): $$\nabla \cdot \vec{E} = \frac{1}{r^2}\frac{d}{dr}(r^2E_r)$$

So we get Gauss's equation as: $$\frac{d}{dr}(r^2E_r) = \frac{q}{\varepsilon_0}r^2\delta_{\mathbb{R}^3}(r\hat{r}) \tag{*}$$

For $r \gt 0$ the equation is just $\frac{d}{dr}(r^2E_r) = 0$, so that $r^2E_r = C$ for some constant $C$.

Therefore $E_r = \frac{C}{r^2}$ for $r \gt 0$.

To find the actual value of $C$, we integrate (*) from $0$ to $\infty$ and use $r^2E_r = C$ to find $$C = \frac{q}{\varepsilon_0}\int_{0}^{\infty}\delta_{\mathbb{R}^3}(r\hat{r})r^2dr$$ (note that this is not fully rigorous since the usual Fundamental theorem of calculus applies to functions, not distributions).

Now the main property of $\delta_{\mathbb{R}^3}(\vec{r})$ is that $$\iiint\limits_{\mathbb{R}^3} \delta_{\mathbb{R}^3}(\vec{r}) dV = 1$$

Using spherical coordinates, this is $$1 = \int\limits_{0}^{\infty}\int\limits_{-\pi}^{\pi}\int\limits_{0}^{2\pi}\delta_{\mathbb{R}^3}(r\hat{r})r^2\sin\varphi \text{ d}\theta\text{ d}\varphi \text{ d}r = \int\limits_{0}^{\infty}\delta_{\mathbb{R}^3}(r\hat{r})r^2\text{ d}r \int\limits_{-\pi}^{\pi}\sin\varphi\text{ d}\varphi\int\limits_{0}^{2\pi}d\theta = 4\pi\int\limits_{0}^{\infty}\delta_{\mathbb{R}^3}(r\hat{r})r^2\text{ d}r$$

Therefore $C = \frac{q}{4\pi\varepsilon_0}$ and we obtain $$\vec{E} = \frac{q}{4\pi\varepsilon_0r^2}\hat{r}$$

On the other hand, this derivation is not actually too different from the one using the integral version of Gauss's theorem, since we performed an integration and used the 1-d fundamental theorem of calculus to find $C$, which is equivalent to using the divergence theorem in 3-d.

  • $\begingroup$ For physicists using Dirac delta functions under integrals like this is rigourous :) $\endgroup$ Oct 13, 2018 at 5:33
  • $\begingroup$ Is it possible not to utilize spherical symmetry? $\endgroup$
    – Szeto
    Oct 13, 2018 at 7:37
  • $\begingroup$ @Szeto Sorry for the late reply. If you dont assume spherical symmetry I think it is still possible to solve it, but you will have to use expansions of $E$ and $\delta_{\mathbb{R}^3}$ in spherical harmonics. Then you can compute the spherical expansion of $\nabla\cdot E$. At this point you will be comparing coefficients in the left and right hand sides, and only the coefficient corresponding to a spherically symmetric term will be nonzero. (We are using the completeness and orthogonality property of spherical harmonics in this argument). $\endgroup$
    – Tob Ernack
    Jun 30, 2019 at 5:36
  • $\begingroup$ @TobErnack If you have time and if you don’t mind, please demonstrate the maths behind. You might want to add it to your answer. Lots of thanks! $\endgroup$
    – Szeto
    Jun 30, 2019 at 5:38

It'd be hard to get the multiplicative factor, I think, but the basic principle is not hard:

The divergence in spherical coordinates is handily provided by Wikipedia as$$\nabla\cdot A = {1 \over r^2}{\partial \left( r^2 A_r \right) \over \partial r} + {1 \over r\sin\theta}{\partial \over \partial \theta} \left( A_\theta\sin\theta \right) + {1 \over r\sin\theta}{\partial A_\varphi \over \partial \varphi}. $$Looking for a spherically symmetric field gives just $$ {d \left( r^2 E_r \right) \over d r} = r^2~ {\rho(r)\over\epsilon_0}. $$ To get the proportionality right one would want to solve for $\rho(r)=\{3q/(4\pi R^3)\text{ if } r< R\text{ else }0\}$ with boundary condition $E_r(r=0)=0.$


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