Suppose I have two regions in space. Region 1 and region 2.

In Region 1 I have a bunch of charges, and in region 2 I have no charges.

Is it true that Laplace's equation is satisfied in region 2?


2 Answers 2


Yes, indeed. The electrostatic potential $V$ satisfies $$\nabla^{2}V\left(\vec{r}\right)=-\frac{\rho\left(\vec{r}\right)}{\varepsilon_{0}}$$

so in a region $U\subset\mathbb{R}^{3}$ in which $\rho=0$, one also has $$\nabla^{2}V\left(\vec{r}\right)=0$$

Note, however, that it doesn't mean you can always do something with it. For example, most theorems on Laplace's equation will require $U$ to be a connected open subset of $\mathbb{R}^{3}$ (with the topology induced by the Euclidean metric). So taking for instance $U$ to be a bunch of discrete points with no charges won't help you so much.


The Poisson equation is a differential equation. Any differential equation on $\Omega\subset \mathbb{R}^n$ can be written in terms of a differential operator $\mathfrak{D}$ mapping functions on $\Omega$ to functions on $\Omega$ as


where $j$ is what sometimes is called a source term. When $j = 0$ the equation is homogenous.

The differential equation is an equality between functions. But equality of functions is pointwise equality. Thus when $f$ is a solution, $\mathfrak{D}f(x) = j(x)$ for all $x\in \Omega$.

In particular if $\Omega'\subset \Omega$ then $\mathfrak{D}f(x)=j(x)$ for all $x\in\Omega'$.

Now if $j(x) = 0$ if $x\in \Omega'$ this means that restricted to $\Omega'$ $f$ will satisfy $\mathfrak{D}f=0$ which is the homogenous equation associated to $\mathfrak{D}$.

Your is just the particular case $\mathfrak{D}=\nabla^2$ with $j(x)=\rho(x)/\epsilon_0$ and with $\Omega'$ the region $2$. The potential is a solution to $\mathfrak{D}\Phi=j$ and thus in $\Omega'$ the potential satisfies Laplace's equation



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