The problem I'm struggling with is a problem from the book "introduction to electrodynamics" by Griffiths.

Here it is:
A time-dependent point charge $q(t)$ is at the origin, $\rho(\mathbf r,t)=q(t)\delta ^3 (\mathbf r)$ is fed by a current $\mathbf J(\mathbf r,t)=-\frac{1}{4\pi}\frac{\dot q} {r^2}\hat r$ and $\dot q=\frac{dq}{dt}$
$\Bbb {a)}$ Check that charge is conserved by confirming that the continuity equation is obeyed.
$\Bbb {b)}$Find the scalar and vector potentials in the Coulomb gauge.
$\Bbb {c)}$ Find the fields and check that they satisfie all of Maxwell's equations.

For $\Bbb a$ I succesfully showed that $\nabla \cdot J =- \frac{d\rho}{dt}$
For $\Bbb b$ I get till the point $V=\frac{q}{4\pi \epsilon_0}\int\frac{\delta ^3}{r}d\tau' $ which is also what my answer model shows, however in the next step it says so $V=\frac{q(t)}{4\pi \epsilon_0 r}$
I don't get how this V is calculated.
Then they say "By symetry $\mathbf B=0$, thus $\mathbf A=0$"
I also don't get wher they get this from.

I didn't yet start on c, since I didn't get parts of b

Help is very much appreciated :)

  • 1
    $\begingroup$ In Coulomb gauge you don't integrate over time to get the potential, you just use Coulomb's law at each instant, hence the name. $\endgroup$
    – Javier
    Commented May 29, 2017 at 20:13

1 Answer 1


As Javier said in the comments, the nice thing about the Coulomb gauge is Poisson's equation and Coulomb's law holds for $V$ and the instantaneous $\rho$ so the answer for $V$ for your part b follows immediately.

The flip side of the Coulomb gauge is the differential equation for $\vec{A}$ is usually uglier, but in this case you can show since $$\vec{J}=\epsilon_0\vec\nabla \left(\frac{\partial V}{\partial t}\right),$$ then $\vec{A}$ just satisfies the wave equation $$\nabla^2 \vec{A}-\mu_0\epsilon_0\frac{\partial^2}{\partial t^2}\vec{A}=0.$$ Now one solution of this equation is indeed $\vec{A}=0$, but it is not the only solution. The other solutions correspond to freely propagating electromagnetic waves, and no symmetry argument eliminates them ($\vec{B}\neq 0$ for these solutions). But in any case these electromagnetic waves are not caused by the changing $\rho$ so it seems this problem wants you to take $\vec A =0.$ You should be able to verify Maxwell's equations whether you take it to vanish or not.


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