How Does Electron Charge Density Affect Electric Field in a Hydrogen Atom?

[bon]

Homework Statement

The electron charge density of a hydrogen atom in its ground state is given by:

p(r) = -e/ pi a^3 exp(-2r/a) where a is the Bohr radius

Show the E field due to the cloud is given by:

E(r) = e/4pi episllon0 ( (exp(-2r/a) -1)/r^2 + 2exp(-2r/a)/ar + 2exp(-2r/a)/a^2)

Homework Equations

The Attempt at a Solution

I know that the E field is given by: http://en.wikipedia.org/wiki/Coulomb\'s_law#Continuous_charge_distribution

I\'m trying to understand a solution I\'ve been provided with... but the solution takes out a factor of 1/4pi e0 r^2 then integrates the expression over all space (r\' ^2 sintheta, dtheta dphi dr\') to get the right answer..

I don't understand why this works...surely each little element of charge is a different distance away from the point r (i.e. r-r\')?

Can anyone explain this to me?

Thanks!

 

[ehild]

In the ground state, the charge distribution has spherical symmetry, independent on direction. It can be assumed the same for the electric field that it is function of r alone. Apply Gauss' Law.

ehild