Find electric potential of field inside and outside nucleus

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SUMMARY

The discussion focuses on deriving the electrostatic potential energy of an electron in the field of a finite nucleus with charge +Ze and radius R = r0A1/3. The derived potential expression is V(r) = -Ze2/r for r > R and V(r) = Ze2/R( (r2/2R2) - 3/2) for r < R. The application of Gauss' law is crucial for determining the electric field inside the nucleus, where the charge density is uniform, leading to the enclosed charge Qenc = Ze(r3/R3). The continuity of potential at R is also emphasized.

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  • Understanding of electrostatics and electric potential
  • Familiarity with Gauss' law
  • Knowledge of Coulomb's law
  • Basic calculus for integration
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  • Study the application of Gauss' law in electrostatics
  • Learn about electric potential and its relationship with electric field
  • Explore the concept of charge density and its implications in electrostatics
  • Investigate continuity conditions for potential in electrostatic systems
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vbrasic
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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.
 
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Same integral as outside the sphere ... :rolleyes:
And at R the potential has to be continuous.
 
BvU said:
Same integral as outside the sphere ... :rolleyes:
And at R the potential has to be continuous.
So essentially all I have to do is first integrate from infinity to R, and then from R to r?
 
I am surprised you have to ask. Yes ! Exactly !
 

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