= Solution
In vacuum SI notation, <Maxwell's equations> are
$$
\nabla\cdot E=\rho/\epsilon_0,\quad\nabla\cdot B=0,\quad
\nabla\times E=-\partial_tB,\quad
\nabla\times B=\mu_0J+\mu_0\epsilon_0\partial_tE.
$$
Taking the <divergence> of the last equation, using that the <divergence> of a <curl> vanishes and then the first equation, gives
$$
0=\mu_0\nabla\cdot J+\mu_0\partial_t\rho,\qquad
\boxed{\partial_t\rho+\nabla\cdot J=0.}
$$
This is local <conservation of electric charge>. For spatially uniform <electrical conductivity>, <Ohm's law> $J=\sigma E$ gives $\nabla\cdot J=\sigma\rho/\epsilon_0$, so
$$
\boxed{\rho(x,t)=\rho(x,0)e^{-\sigma t/\epsilon_0},\qquad\text{decay rate }\sigma/\epsilon_0.}
$$
For a homogeneous dielectric conductor with permittivity $\epsilon$, replace $\epsilon_0$ by $\epsilon$. This <charge relaxation> is a statement about bulk charge; charge transported to a boundary is still accounted for by the conservation law.
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