Uniformly charged sphere's electric field I am facing this topic for the umpteenth time in my college career and, of course, every teacher has explained it in a different way.
In this course, to find the expression of the electric field of a uniformly charged sphere, the professor started from this equation.
$$
\frac{1}{r^2}\ \frac{\partial}{\partial r}\  \lgroup r^2 \frac{\partial u}{\partial r}\  \rgroup = -\frac{q}{\epsilon}\ \delta(r) 
$$
some idea of ​​what he is doing?
 A: From Gauss's law & using the conservative field, we have
$$
\nabla^2u=-\frac\rho{\varepsilon_0}
$$
In spherical coordinates and assuming isotropy in $\theta$ and $\phi$ directions, the above becomes,
$$
\frac{1}{r^2}\frac{\partial}{\partial r}\left(r^2\frac{\partial u}{\partial r}\right)=-\frac{\rho}{\varepsilon_0}\tag{1}
$$
Since we are dealing with a point-charge, the charge density can be expressed in terms of the Dirac delta function:
$$
\rho(\mathbf r)=q\delta(\mathbf r-\mathbf r')=q\delta(\mathbf r)\tag{2}
$$
where we assume that $\mathbf r'=O$ (the origin) and ignore it in the delta function. This can be done because the volume integral of $\rho$ should give the total charge $q$:
$$
\int\rho\,dV=\int q\delta(\mathbf r)\,dV=q\int\delta(\mathbf r)\,dV=q
$$
where use used the property $\int\delta(x-x')\,dx=1$ when $x'$ is in the integral range.
Inserting (2) into (1), we get
$$
\frac{1}{r^2}\frac{\partial}{\partial r}\left(r^2\frac{\partial u}{\partial r}\right)=-\frac{q}{\varepsilon_0}\delta(\mathbf r)
$$
