# Electrostatic Energy in the Hydrogen Atom

## Homework Statement

We model the Hydrogen atom as a charge distribution in which the proton (a point charge) is surrounded by negative charge with the volume density of ρ = -ρ0 * exp (-2r/a0) where a0 is the Bohr radius. And ρ0 is a constant chosen such that the entire atomic distribution is neutral. What is the electrostatic energy contained within the electric field of outer region of the sphere of radius a0?

## The Attempt at a Solution

I have tried reducing the sphere to a point charge, but it doesn't work for neutral distributions (Should've figured that sooner.) Generally, I've tried to apply everything I could as far as the formula W (energy) = ∫ φ * dq goes (where φ is the electrostatic potential, and dq is the charge.)

Homework Helper
Gold Member
I think you need to build up the charge density around the proton as a function of radius and as you begin to add layers of negative charge, the electric field and the potential will be reduced in amplitude. Besides the negative electrostatic energy from the negative electric charge distribution, the electric field energy density has a term that i believe is ## U=E^2/(8 \pi) ## in cg.s. units. The question seems to be asking for the electric field energy for ## r>a_o ##, but the question doesn't appear to be entirely clear. The electric field can be computed as a function of r using Gauss's law. The question being asked seems to need some additional clarification. By electrostatic energy, do they mean electric charge distribution interacting with electric field or just the electric field energy?

I reckon merely the electric field energy. Though, unfortunately, that is as clear as the question gets...For a superficial distribution I was able to simply assess the potential of interaction between the proton and the electron, and also the potential the electron generates when it interacts with itself, then I used said potential to calculate the energy as the sum between the two respective interaction energies. Though it seems that making it a volume distribution rather than a surface one complicates things further.