# Applications of Gauss's Law

1. Feb 18, 2006

### Brewer

First could someone actually explain the following question to me!

Consider a hydrogen atom to be a positive point charge e at the center of a uniformly charged sphere of radius R and with total charge -e. Using Gauss' Law, find an expression for the electric field as a function of the distance r from the nucleus. [Hint: Determine the fields of the nucleus and electron separately, then the total field]

As far as I can understand from my notes

flux = Q/permittivity of free space.

so that would mean for the positive charge would be 1.81*10^-8 Nm^2/C.

But after this it no longer makes sense, because one part of my notes says it doesn't depend on the radius of the sphere, but as I read the question, it obviously does.

I'm so confused!! Could someone please help me out before I go nuts!

Thanks guys

2. Feb 18, 2006

### Staff: Mentor

Consider what Gauss' law tells you: The total flux through any closed surface (for example, a sphere of radius r) will equal $Q/\epsilon_0$, where Q is the total charge contained within that surface. Express that statement mathematically and see where it takes you.

3. Feb 18, 2006

### Brewer

I don't want to sound totally thick, but huh? I understand that about flux (I think), but where does it take into account the distances between them.

As I can see, the proton in the center of the neutron (the positive charge) will at a certain distance from it have a Gaussian surface or some degree or another. But won't the electron flying around the outside have that as well? So at some point because they are the same magnitude of charge, won't they cancel each other out? Gauss' Law isn't used to work out field strength is it?

Its such a confusing question...

4. Feb 18, 2006

### nrqed

I wil get you started and then you can write back and ask more. I assume you know how to find the E field of the proton at the center, that`s just the E field of a point charge.

For the electron cloud, the charge is uniformly spread out over a sphere of radius R. Nowyou must use Gauss' law applied to a point at a distance r <R to find the E field for r<R. So your gaussian surface is a sphere of radius r < R. It has a surface area 4 pi r^2 and a volume 4/3 Pi r^3. Its charge density is the total charge of the electric cloud divided by the total volume of the electric cloud, so -e/(4/3 Pi R^3).

(Sorry.. I am on a French keyboard and can't find the symbols I need to type in tex stuff)

The electric flux in that region is - 4 pi r^2 E (where E is the magnitude of the E field and the reason for the minus sign is that the E field is pointing toward the origin. Have you seen how to calculate flux as an integral over a surface?).

Now Gauss' law states that the flux = Q_enclosed/epsilon_0
And Q_enclosed is the charge density of the cloud times the volume of the gaussian surface 9all given above).

Plugging everything into Gauss' law will give you the E field produced by the electron (its magnitude...its direction will be toward the origin)

hope this helps

Pat

5. Feb 18, 2006

### Staff: Mentor

In this problem you are asked to model the electron, not as a flying point charge, but as a uniform distribution of charge filling a sphere of radius R (and adding up to a charge of -e). So the amount of negative charge contained within any Gaussian surface will certainly depend on its distance r from the proton.

6. Feb 19, 2006

### Brewer

I think that it helps. We shall have to see. Am I likely to get an R in the answer. As far as I can see I have got one. In fact I have one cubed (I can't see how to cancel it, but I have the feeling it should). However most of that made quite a lot of sense.

Thanks for the assistance.

7. Feb 19, 2006

### Staff: Mentor

Why does having an R in the answer bother you? (It's just a constant.) For r < R, the field will depend on R; for r > R, it won't.

8. Feb 19, 2006

### nrqed

As someone else said, yes, your answer will indeed contain R. And you are right, it contains R^3 (because the charge density contains the total volume of the charge cloud). So you are on the right track.

Pat

9. Feb 26, 2006

### Brewer

Well that sounds promising then! Hopefully I have my results back tomorrow!

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