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## Homework Statement

Derive following expression for the electrostatic potential energy of an electron in the field of a finite nucleus of charge, ##+Ze##, and radius, ##R=r_0A^{1/3}##, where ##r_0## is a constant. (Charge density is constant.)

The potential we are asked to derive is:

$$

V(r) = \begin{cases}

\frac{-Ze^2}{r} & \text{if } r>R\\

\frac{Ze^2}{R}(\frac{r^2}{2R^2}-\frac{3}{2}) & \text{if } r<R

\end{cases}.

$$

## Homework Equations

Gauss' law.

## The Attempt at a Solution

Naturally, for ##r>R##, we approximate the nucleus as being point-like, with electric field of magnitude, $$\frac{Ze}{4\pi\epsilon_0r^2}.$$ I'm not sure how to find potential, though physical intuition suggests it's just the standard Coulomb potential.

However, inside the nucleus we use Gauss' law. We have that the charge enclosed is ##Q_{enc}=Ze\frac{r^3}{R^3}##, for a uniform charge density. Then using Gauss' law, we have, $$\int |E|da=Ze\frac{r^3}{R^3\epsilon_0}\rightarrow |E|=Ze\frac{r}{4\pi\epsilon_0R^3}$$ directed radially, such that, $$E=Ze\frac{r}{4\pi\epsilon_0R^3}\hat{r}.$$ I'm not sure how to get potential from here, or exactly what integral I'm supposed to use to get the derived potential above.