Homework Help: Calculate electric field produced by total charge density

1. Sep 19, 2013

wifi

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 $$\rho_n(\vec{r})=2q\delta^3(\vec{r})$$ and $$\rho_e(\vec{r})=-\frac{8q}{\pi {a_0}^3}e^{-4r/{a_0}}$$

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

Attempt at a Solution:

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

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?

2. Sep 19, 2013

TSny

The electron cloud of one of the electrons extends all the way to infinity. So, if your Gaussian surface has a finite radius, it will not enclose all of the charge of the electron.

So, think about how you can find the charge inside the Gaussian surface.

3. Sep 19, 2013

wifi

Wouldn't the only way to enclose all of the charge of the electron be to have a sphere with an infinite radius act as our Gaussian surface?

4. Sep 19, 2013

TSny

Why do you want to enclose all of the charge? To find the electric field at a distance $r$ from the nucleus, you'll need a Gaussian surface of radius $r$. So, you're going to have to find the electric charge enclosed inside that Gaussian surface.

5. Sep 19, 2013

wifi

Oh right! So I'll wind up with an equation for the electric field $\vec{E}$ as a function of $r$.

So I start with the LHS of Gauss's law: $\oint \vec{E} \cdot d\vec{A}$

$\vec{E}$ and $d\vec{A}$ will be parallel, so we get $\int |\vec{E}|d\vec{A}$. And since the magnitude of $\vec{E}$ will be constant over the surface, we have $|\vec{E}| \int d\vec{a}=|\vec{E}|4\pi r^2$.

Comparing to the RHS:

$|\vec{E}|4\pi r^2=\frac{Q_{enc}}{\epsilon_0}$

So wouldn't the enclosed charge be $2q$ (due to the nucleus) plus whatever fraction of the electron cloud is enclosed in the Gaussian surface?

EDIT: Left out $|\vec{E}|$

Last edited: Sep 19, 2013
6. Sep 19, 2013

TSny

Yes, that's right. (I think you left out E in the last equation.)

7. Sep 19, 2013

wifi

Okay thanks!

So I'm getting $\vec{E}=\frac{1}{4\pi \epsilon_0}\frac{q}{r^2}\hat{r}$ as my final answer.

(Fixed the error in my previous post as well.)

EDIT: Oh wait... pretty sure this is wrong because I didn't account for the electron cloud only partially contributing to the enclosed charge. So should it be $Q_{enc}=2q+\int_0^r \rho_e(\vec{r}) dr$?

8. Sep 19, 2013

TSny

No, that's not the right answer.

How did you find $Q_{enc}$ (the net charge inside the Guassian surface of radius r)?

9. Sep 19, 2013

wifi

I don't know if you saw the edit to my last post, but I realized it was wrong right after I posted it. But the charge enclosed in the Gaussian surface should be $Q_{enc}=2q+\int_0^r \rho_e(\vec{r}) dr$, I believe.

10. Sep 19, 2013

TSny

Almost. Note that $\rho_e(r)$ is a volume charge density. So, you should integrate over the volume enclosed by the Gaussian surface.

11. Sep 19, 2013

wifi

Darn, so close.

So then I have $Q_{enc}=2q+\int_0^r \rho_e(\vec{r})d\tau$.

And the integral is then $\int_0^r \rho_e(\vec{r})d\tau=\int_0^{2 \pi} \int_0^{\pi} \int_0^{r}(-\frac{8q}{\pi {a_0}^3}e^{-4r/{a_0}})r^2sin\phi \ dr \ d\theta \ d\phi$.

12. Sep 19, 2013

TSny

Yes, that's it.

13. Sep 19, 2013

wifi

One last thing:

$2q$ comes from the fact that $$\int_0^r \rho_n (\vec{r}) d\tau=\int \int \int 2q \delta^3(\vec{r})r^2 sin\phi \ dr \ d\theta \ d\phi=2q \int \int \int \delta^3(\vec{r})r^2 sin\phi \ dr \ d\theta \ d\phi = 2q \cdot 1 = 2q$$ Correct?

14. Sep 19, 2013

TSny

Essentially, yes. Working with the the Dirac delta function in spherical coordinates is a little tricky.

See for example equation 51 here: http://mathworld.wolfram.com/DeltaFunction.html Note that in spherical coordinates your integration over r starts at r = 0, so the integration region doesn't really "contain" r = 0 (the nucleus) as an interior point in the region of integration. That makes me a little nervous.

To me, it's easier to think of going to Cartesian coordinates for taking care of the delta function at the origin

$\delta^3(\vec{r}) = \delta(x)\delta(y)\delta(z)$.

But stick with spherical coordinates for the integration of the electron charge distribution.

15. Sep 20, 2013

vanhees71

That's correct. Spherical coordinates are singular at $\vec{r}=0$ (even along the entire polar axis), and thus you cannot express properly $\delta^{(3)}(\vec{r})$ in terms of spherical coordinates.