An early model for an atom considered it to have a positively charged point nucleus of charge $Ze$,surrounded by a uniform density of negative charge up to a radius $R$. The atom as a whole is neutral. For this model,what is the electric field at a distance $r$ from the nucleus?

(N/A) Solution:
The charge distribution for this model of the atom is as shown in the figure. The total negative charge in the uniform spherical charge distribution of radius $R$ must be $-Ze$,since the atom (nucleus of charge $Ze$ + negative charge) is neutral. This immediately gives us the negative charge density $\rho$,since we must have:
$\frac{4}{3} \pi R^{3} \rho = -Ze$
$\rho = -\frac{3Ze}{4 \pi R^{3}}$
To find the electric field $E(r)$ at a point $P$ which is a distance $r$ away from the nucleus,we use Gauss's law. Because of the spherical symmetry of the charge distribution,the magnitude of the electric field $E(r)$ depends only on the radial distance $r$. Its direction is along the radius vector from the origin to the point $P$. The Gaussian surface is a spherical surface centered at the nucleus. We consider two situations: $r < R$ and $r > R$.
$(i)$ $r < R$: The electric flux $\phi$ enclosed by the spherical surface is $\phi = E(r) \times 4 \pi r^{2}$. The charge $q$ enclosed by the Gaussian surface is the positive nuclear charge and the negative charge within the sphere of radius $r$,i.e.,$q = Ze + \frac{4}{3} \pi r^{3} \rho$. Substituting for the charge density $\rho$,we have $q = Ze - Ze \frac{r^{3}}{R^{3}}$. Gauss's law gives $E(r) \times 4 \pi r^{2} = \frac{q}{\varepsilon_{0}}$,so $E(r) = \frac{Ze}{4 \pi \varepsilon_{0}} \left( \frac{1}{r^{2}} - \frac{r}{R^{3}} \right)$. The electric field is directed radially outward.
$(ii)$ $r > R$: In this case,the total charge enclosed by the Gaussian spherical surface is zero since the atom is neutral. Thus,from Gauss's law,$E(r) \times 4 \pi r^{2} = 0$,which implies $E(r) = 0$ for $r > R$.