# Gauss's law, nucleus, potential proton due to other protons

1. May 5, 2015

### binbagsss

I'm a little out of touch with this stuff , but I'm really not getting it..

So my book is considering: in a nucleus of $z$ protons, consider one proton in spherical charge distribution to other protons.

So $\rho = (z-1)e/(4/3) \pi R^{3}$, where $R$ is the radius of the nucleus, is the charge density. (this is fine)

I have for Gauss's law:

$\int E.d\vec{A} = Q_{enclosed}/\Sigma_{0}$,

So I get $E.4\pi r^{2}=\rho (4/3) \pi r^{3} / \Sigma_{0}$
$=>$ $E=(z-1)e r/4\pi R^{3}$

Whereas my book has $q/4 \pi \Sigma_{0} r= (4/3) \pi r^{3} \rho/4 \pi \Sigma_{0}r=(z-1)e/4 \pi \Sigma_{0}(r^{2}/R^{3}$,

I dont understand the far LHS of this? But anyway I'm a factor $r$ out, I have no idea how the book gets $r^{2}$,

your assistance is greatly appreciated, thank you !

Last edited: May 5, 2015
2. May 5, 2015

### BvU

( I hope it's OK with you if I use $\varepsilon_o$ for the permittivity of the vacuum. With your symbol, I keep wondering what gets summed )

$${1\over 4 \pi \varepsilon_o}\; {q\over r}$$ looks more like a potential, not an electric field strength. That would explain the factor r.