= Solution
<Gauss's law> states that the outward <electric flux> through a closed <surface> is the enclosed <electric charge> divided by the <vacuum permittivity>:
$$
\oint_S\mathbf E\cdot d\mathbf S=\frac{Q_{\rm enc}}{\epsilon_0},\qquad \nabla\cdot\mathbf E=\frac{\rho}{\epsilon_0}.
$$
In <electrostatics>, the <electric field> is the negative <gradient> of the <electrostatic potential>, $\mathbf E=-\nabla V$. Substitution into the <differential equation> form of <Gauss's law> gives <Poisson's equation> $\nabla^2V=-\rho/\epsilon_0$. Consequently, in a region containing no <charge density>,
$$
\boxed{\nabla^2V=0,}
$$
which is <Laplace's equation>. Charges on the bounding <conductors> enter through the <boundary conditions>; they need not vanish for <Laplace's equation> to hold inside the empty gap.
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