= Solution
Use SI units and the <Minkowski metric> $\eta=\operatorname{diag}(1,-1,-1,-1)$. Write $x^\mu=(ct,x,y,z)$, $\partial_\mu=(c^{-1}\partial_t,\nabla)$ and the <four-current> $J^\mu=(c\rho,\boldsymbol J)$, with Greek indices from $0$ to $3$. The <electromagnetic field tensor> is the antisymmetric tensor whose components are
$$
F^{0i}=-E_i/c,\qquad F^{i0}=E_i/c,\qquad F^{ij}=-\epsilon_{ijk}B_k.
$$
Indices are raised and lowered with $\eta$. Define the <dual electromagnetic field tensor> by $\widetilde F^{\mu\nu}=\tfrac12\epsilon^{\mu\nu\rho\sigma}F_{\rho\sigma}$, with $\epsilon^{0123}=+1$. Then \b[the covariant <Maxwell equations> are]
$$
\boxed{\partial_\mu F^{\mu\nu}=\mu_0J^\nu,\qquad
\partial_\mu\widetilde F^{\mu\nu}=0.}
$$
Here $\rho$ and $\boldsymbol J$ are charge and current density, $\mu_0$ is the vacuum <magnetic permeability>, and $\epsilon_0\mu_0c^2=1$. For example the $\nu=0$ equation gives $\nabla\cdot\boldsymbol E=\rho/\epsilon_0$, and the spatial equations give $\nabla\times\boldsymbol B-c^{-2}\partial_t\boldsymbol E=\mu_0\boldsymbol J$. With the stated dual convention, $\widetilde F^{0i}=-B_i$ and $\widetilde F^{ij}=\epsilon_{ijk}E_k/c$; the second tensor equation gives the two homogeneous <Maxwell equations>.
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