Newtonian Potential satisfies Laplace's Equation

Theorem
Let $R$ be a region of space.

Let $S$ be a Newtonian potential over $R$ defined as:
 * $\forall \mathbf r = x \mathbf i + y \mathbf j + z \mathbf k \in R: \map S {\mathbf r} = \dfrac k r$

where:
 * $\tuple {\mathbf i, \mathbf j, \mathbf k}$ is the standard ordered basis on $R$
 * $\mathbf r = x \mathbf i + y \mathbf j + z \mathbf k$ is the position vector of an arbitrary point in $R$ with respect to the origin
 * $r = \norm {\mathbf r}$ is the magnitude of $\mathbf r$
 * $k$ is some predetermined constant.

Then $S$ satisfies Laplace's equation:
 * $\nabla^2 S = 0$

where $\nabla^2$ denotes the Laplacian.

Proof
From Gradient of Newtonian Potential:


 * $\grad S = -\dfrac {k \mathbf r} {r^3}$

where $\grad$ denotes the gradient operator.

Then: