Spherically Symmetric Charge Distribution

[steelclam]

Homework Statement

Consider a spherically symmetric charge distribution [tex] \rho = \rho (r) [/tex]

Homework Equations

By dividing the charge distribution into spherical shells, find the potential [tex] \phi [/tex] and the electric field strength [tex] \bf{E} [/tex] in terms of [tex] \rho (r) [/tex]

The Attempt at a Solution

The given solution is

[tex] \phi (r) = \frac{4 \pi}{r} \int_0^r \rho (r\prime) {r \prime}^2 dr \prime + 4\pi \int_r^\infty \rho (r\prime ) r \prime dr \prime
[/tex]
I just can't understand the separation in two integrals, I think it is not rigorous...by the way the book (Problems in Electrodynamics, Batygin and Toptygin) uses non-SI units.

 

[Nate810]

A:The two integrals are just a consequence of the definition of the electric potential. By definition, the electric potential at a point $P$ due to a point charge $q$ is $\phi = \frac{1}{4\pi \epsilon_0} \frac{q}{r}$, where $r$ is the distance between the point charges. Now, when you have an extended charge distribution (as in your case), we have to add the contributions of all the little pieces of charge that are making up the whole charge distribution. This is done by integrating over all the charge elements, i.e. $$\phi(P) = \frac{1}{4 \pi \epsilon_0} \int_{V'} \frac{\rho(r')}{|\vec{r}-\vec{r'}|} dV'$$where the integral is taken over the volume $V'$ containing all the charge elements. In your case, it turns out that the charge distribution is spherically symmetric, and so there is no dependence on the angles in the above integral, and it can be written as$$\phi(P) = \frac{1}{4 \pi \epsilon_0} \int_0^\infty \frac{\rho(r')}{|\vec{r}-\vec{r'}|} 4 \pi r'^2 dr'$$where the integral limits have been changed to go from 0 to $\infty$ since we have considered a spherically symmetric charge distribution. Now, the separate integrals in your solution can be understood if you first split the range of the integral into two parts and then recombine them. That's what has been done in your solution.