= Solution
Taking the <divergence> of the <Ampère-Maxwell equation> and using the <Gauss's law> $\nabla\cdot E=\rho/\varepsilon_0$ gives
$$
0=\mu_0\nabla\cdot J+\mu_0\frac{\partial\rho}{\partial t},
$$
which is <charge conservation from Maxwell equations> $\rho_t+\nabla\cdot J=0$.
By <Ohm's law>, $J=\sigma E$, so $\nabla\cdot J=\sigma\rho/\varepsilon_0$. Therefore
$$
\boxed{\rho(t)=\rho(0)e^{-\sigma t/\varepsilon_0},}
$$
with charge-relaxation rate $\sigma/\varepsilon_0$ and time $\varepsilon_0/\sigma$.
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