Editing Laplace operator (section)

=== Density associated with a potential ===
If {{math|''φ''}} denotes the [[electrostatic potential]] associated to a [[charge distribution]] {{math|''q''}}, then the charge distribution itself is given by the negative of the Laplacian of {{math|''φ''}}:
<math display="block">q = -\varepsilon_0 \Delta\varphi,</math>
where {{math|''ε''<sub>0</sub>}} is the [[electric constant]].

This is a consequence of [[Gauss's law]]. Indeed, if {{math|''V''}} is any smooth region with boundary {{math|∂''V''}}, then by Gauss's law the flux of the electrostatic field {{math|'''E'''}} across the boundary is proportional to the charge enclosed:
<math display="block">\int_{\partial V} \mathbf{E}\cdot \mathbf{n}\, dS = \int_V \operatorname{div}\mathbf{E}\,dV=\frac1{\varepsilon_0}\int_V q\,dV.</math>
where the first equality is due to the [[divergence theorem]]. Since the electrostatic field is the (negative) gradient of the potential, this gives:
<math display="block">-\int_V \operatorname{div}(\operatorname{grad}\varphi)\,dV = \frac1{\varepsilon_0} \int_V q\,dV.</math>

Since this holds for all regions {{mvar|V}}, we must have
<math display="block">\operatorname{div}(\operatorname{grad}\varphi) = -\frac 1 {\varepsilon_0}q</math>

The same approach implies that the negative of the Laplacian of the [[gravitational potential]] is the [[mass distribution]]. Often the charge (or mass) distribution are given, and the associated potential is unknown. Finding the potential function subject to suitable boundary conditions is equivalent to solving [[Poisson's equation]].