Example

Question:

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?

Solution:

The total negative charge in the uniform spherical charge distribution of radius \( R \) must be \( -Ze \). Hence the negative charge density \( \rho \) is \[ 4\pi R^3 / 3 \cdot \rho = -Ze \implies \rho = -\frac{3Ze}{4\pi R^3} \] To find the electric field \( \mathbf{E}(r) \) at a point \( P \) a distance \( r \) from the nucleus, use Gauss's law.
The charge \( q \) enclosed by a spherical Gaussian surface of radius \( r \) is \[ q = Ze + \frac{4\pi r^3}{3} \rho \] Substitute \( \rho \): \[ q = Ze - Ze \frac{r^3}{R^3} \] By Gauss's law, \[ E(r) = \frac{Ze}{4\pi\varepsilon_0} \frac{1}{r^2} \left(1 - \frac{r^3}{R^3} \right), \quad \text{for } r < R \] For \( r > R \), the total enclosed charge is zero (since the atom is neutral), so \[ E(r) = 0 \quad \text{for } r > R \] At \( r = R \), both formulas yield \( E = 0 \).