# Calculate electric field produced by total charge density

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?

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TSny
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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.

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.
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?

TSny
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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.

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

Yes, that's right. (I think you left out E in the last equation.)
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 $Q_{enc}=2q+\int_0^r \rho_e(\vec{r}) dr$?

TSny
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No, that's not the right answer.

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

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.

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

Darn, so close.

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

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

TSny
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Yes, that's it.

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?

TSny
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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.

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.