- #1

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**Problem:**

In a neutral He atom consisting of a positively charged nucleus of charge ##2q## and two electrons each of charge ##-q##, the volume charge density for the nucleus and for the single electron cloud are respectively given by [tex]\rho_n(\vec{r})=2q\delta^3(\vec{r})[/tex] and [tex] \rho_e(\vec{r})=-\frac{8q}{\pi {a_0}^3}e^{-4r/{a_0}}[/tex]

Using Gauss's law, calculate the electric field ##\vec{E}## produced by the total charge density [tex]\rho(\vec{r})=\rho_n(\vec{r}) + \rho_e(\vec{r})[/tex].

**Attempt at a Solution:**

I'm sure I'll be required to use the integral form of Gauss's law, namely [itex]\oint \vec{E} \cdot d\vec{A}=\frac{Q_{enc}}{\epsilon_0}[/itex].

It's easy to show that the total charge of the electron cloud is equal to ##-q##. Thus, since the nucleus contributes ##2q## and we have, ##Q_{enc}=2q-q=q##; that is, if we have construct a spherical Gaussian surface enclosing the nucleus and one of the electrons. Is this the correct approach?