Answer by Ron Maimon for Laplacian of $1/r^2$ (context: electromagnetism and...
The electric field from your potential is:$$E(r) = {2\over r^3}$$Using Gauss's law, the total charge in a sphere of radius R is:$$Q(r) = \oint E \cdot dS = 4\pi r^2 {2\over r^3} = {8\pi\over r}$$The...
View ArticleAnswer by Qmechanic for Laplacian of $1/r^2$ (context: electromagnetism and...
I) Problem. We put (four pi times) the permittivity $4\pi\varepsilon=1$ equal to one (in the SI unit system) from now on for simplicity. Let us first rephrase OP's question a bit. Instead of starting...
View ArticleAnswer by yayu for Laplacian of $1/r^2$ (context: electromagnetism and...
Vladimir's answer is off by factor of 2. The laplacian is $\nabla^2(\frac{1}{r^2}) = \frac{4}{r^4}$ A potential that falls of as $\frac{1}{r^2}$ is a dipole (In general, if it falls off as $r^{-n}$ its...
View ArticleAnswer by Hans de Vries for Laplacian of $1/r^2$ (context: electromagnetism...
The operator to go from one potential to the other is.$-\frac{\partial }{\partial r}\left\{\frac{q}{4\pi r}\right\} ~~=~~ \frac{q}{4\pi r^2}$and therefor the source in the center which is given by the...
View ArticleAnswer by Vladimir Kalitvianski for Laplacian of $1/r^2$ (context:...
Let us see: $\Delta \frac{q}{r^2} = \frac{1}{r^2}\frac{\partial}{\partial r} \left( r^2\frac{\partial} {\partial r}\frac{q}{r^2}\right)=\frac{2q}{r^4}$. Thus the charge density $\rho(r)$ is...
View ArticleLaplacian of $1/r^2$ (context: electromagnetism and Poisson equation)
We know that a point charge $q$ located at the origin $r=0$ produces a potential $\sim \frac{q}{r}$, and this is consistent with the fact that the Laplacian of $\frac{q}{r}$ is...
View Article