Helmholtz's theorem and charge density

[Ahmed1029]

According to Helmholtz’s theorem, if electric charge density goes to to zero as r goes to infinity faster than 1/r^2 I'm able to construct an electrostatic potential function using the usual integral over the source, yet I don't understand how this applies to a chunk of charge in some region of space, regarding how fast its charge density does to zero. For instance, suppose I only have a small sphere of charge, how fast does does charge density go to zero as r goes to infinity?

 

[vanhees71]

If you have a charge density different from 0 only within a finite region in space (mathematically speaking, $\rho$ has "compact support"), the assumptions of the theorem are fulfilled, because it's strictly 0 for $r=\vec{x}>R$ for some radius, $R$. Then the solution of the Poisson equation (in SI units),
$$\Delta \Phi(\vec{x})=-\frac{1}{\epsilon_0} \rho(\vec{x})$$
is given by "summing up Coulomb potentials", i.e.,
$$\Phi(\vec{x})=\int_{B_R} \mathrm{d}^3 x' \frac{\rho(\vec{x}')}{4 \pi \epsilon_0 |\vec{x}-\vec{x}'|},$$
where $B_R$ is the solid sphere ("ball") of radius $R$ around the origin, which by definition contains all the charge there is.

Helmholtz's theorem in this form is for sure also valid under the conditions you quote, because if $\rho(\vec{x}) = \mathcal{O}(r^{-\alpha})$ for $|\vec{x}|\rightarrow \infty$ and $\alpha>2$, then
$$\mathrm{d}^3 x' \frac{\rho(\vec{x}')}{|\vec{x}-\vec{x}'|} = \mathrm{d} r' \mathrm{d\vartheta'} \mathrm{d} \varphi' \sin \vartheta' \mathcal{O}(r^{\prime -\alpha+1}),$$
i.e., the integrand falls faster with $r'$ than $1/r$, and thus the integral over $r'$ converges for $r' \rightarrow \infty$. The integrals over the angles is over finite regions ($\vartheta \in (0,\pi)$, $\varphi \in (0,2 \pi)$) and thus don't diverge.

The theorem even works if $\rho=\mathcal{O}(1/r^\beta)$ with $\beta>1$, because the potential is defined only up to an additive constant anyway. In this case you just subtract a clever constant, defining
$$\Phi(\vec{x})=\int_{\mathbb{R}^3} \mathrm{d}^3 x' \frac{\rho(\vec{x}')}{4 \pi \epsilon_0} \left (\frac{1}{|\vec{x}-\vec{x}|}-\frac{1}{|\vec{x}_0-\vec{x}'|} \right).$$
The expression in the parentheses goes like $\mathcal{O}(r^{\prime 2})$ for $r' \rightarrow \infty$, and thus the integral converges even under the more general condition on $\rho$.

 

[Ahmed1029]

got it thanks