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Find electric potential of field inside and outside nucleus

  1. Mar 4, 2017 #1
    1. The problem statement, all variables and given/known data
    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:
    $$
    V(r) = \begin{cases}
    \frac{-Ze^2}{r} & \text{if } r>R\\
    \frac{Ze^2}{R}(\frac{r^2}{2R^2}-\frac{3}{2}) & \text{if } r<R
    \end{cases}.
    $$

    2. Relevant equations
    Gauss' law.

    3. 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.
     
  2. jcsd
  3. Mar 4, 2017 #2

    BvU

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    Same integral as outside the sphere ... :rolleyes:
    And at R the potential has to be continuous.
     
  4. Mar 4, 2017 #3
    So essentially all I have to do is first integrate from infinity to R, and then from R to r?
     
  5. Mar 5, 2017 #4

    BvU

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    I am surprised you have to ask. Yes ! Exactly !
     
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