# B The Repulsive Interaction between 2 electrons

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1. Feb 19, 2017

### Karim Habashy

Hi All,

My question goes as follows:

Suppose that, we charge a conducting very thin spherical shell in 'empty space' with a charge equivalent to 16 electrons. The radius (R) of the shell is 20 light years. We wait 20+ years for the electrons to reach equilibrium. Then, we approach one electron on the shell (lets assume one knows the exact position of this electron) while having a shielded electron (q) in hand and at a distance of 1nm we set the electron free from its shielding, so that both electrons now are 1nm apart (at what we can call time = 0).
So the question is, what is the force on the free electron at that time ?, according to the electrostatics of a charged spherical shell its E(R)*q, where E(R) = 16*q/4πε*R^2.

Last edited: Feb 19, 2017
2. Feb 19, 2017

### BvU

According to the Gauss theorem you can't just 'shield an electron' without doing all kinds of things that disturb your scenario.
Apart from that, at time 0 you can ignore the response of the other 15 electrons that will take tens of years to reach the scene where the change occurs

3. Feb 19, 2017

### Ibix

There are something like 1055 atoms in a one atom thick spherical shell that size. If it's a conductor, each one has at least one conduction electron free to move. I don't think your extra sixteen electrons will stand out at all.

4. Feb 19, 2017

### BvU

My perception was that he shell is empty except for the 16 electrons. @Karim Habashy ?

5. Feb 19, 2017

### Karim Habashy

@BvU my assumption is something of this sort, or that we can ignore the contributions from the atoms and their electronic clouds.
The shell is initially electrically neutral and charging the shell with any charge (Q) will give a charge denisty Q/4πR^2.
Also, lets look at the other extreme, we can charge the shell with 10^55 electrons, so as to have an extra/electron per atom. So will the force on a 1nm away electon be q*(q*10^55)/4πε*R^2 ? (I think its easier to think of smaller number of electrons).
Also, I can approach the shell with an electron and positron so close together so as to minimize the initial intraction, and then i quickly move the positron far away. Or, can we just assume that the shell and the electron exist in this configuration (some how at time = 0).

6. Feb 19, 2017

### Ibix

In a conductor, the atoms have an electron cloud each - but there is also a sea of conduction electrons that are free to move around the shell to even out the charge. That's how conductors do what they do. So I think you'll have real trouble localising any extra electron.

If you simply arrange sixteen charges in empty space, that's a different matter.

7. Feb 19, 2017

### Ibix

Agree that speed of light limits matter - but I think that just means that the immediate reaction looks like an very very weakly charged infinite plane.

8. Feb 20, 2017

### Karim Habashy

The argument includes waiting 20+ or more for equilbrium to be established, so that the information about the presence of the 10^55 electrons is comunicated across all space.

9. Feb 20, 2017

### BvU

Karim, does you scenario have to be so complicated ? Can't we just have 2 electrons far apart and bring in a third one from infinity ?

10. Feb 20, 2017

### Karim Habashy

BvU, I think my scenario is quite simple. I will try to make it simpler to emphasize my point and I will start with a picture:
http://[ATTACH=full]200201[/ATTACH]
So what is the Electric field at q? given that the sphere has symmetric charge distripution of ex.14 (I have only drawn in 2D). Is the field 12*q/4πε*R^2 (according to Gauss) or q/4πε*(10nm)^2.

Last edited by a moderator: May 8, 2017
11. Feb 20, 2017

### BvU

Much clearer ! You have no symmetry under small rotations, so the Gauss result isn't valid.

12. Feb 20, 2017

### Karim Habashy

How small is small rotations ?

13. Feb 20, 2017

### BvU

$\pi\over 8$ in the picture, for instance changes distance to nearest elecron from 10 nm to $5\pi$ light years. That's quite a change ...

14. Feb 20, 2017

### Karim Habashy

Ok we can take the argument to the other extreme, We can charge the sphere unifromally with 10^48 electrons to have have about 0.4 electrons /nm2, will the E-field at q be (10^48)*q/4πε*R^2 ?. The point is (I think) we can charge the sphere with any number of point charges and most of the time we won't get the right field by Gauss's law.

15. Feb 20, 2017

### BvU

$4\pi r^2 = 4.5 \times 10^{35}$ m2 for 20 ly, so with $10^{48}$ electrons you don't get much more than two electrons per $\mu m^2$.

Need a factor 1 million more electrons to reduce that to nm2 and by then their separation is such that the field at 10 nm is as good as radial and Gauss flies.

16. Feb 20, 2017

### Karim Habashy

I think I made a mistake in the input of the light year value, thanks for the correction. So, can we say that Gauss's law is limited to the charge distributions which are very dense (relative to the distance to the outside point of field application) so as to have a uniform radial electric field at that point?.

17. Feb 20, 2017

### BvU

Yes. Note that 10 nm is really very small.
And effective use of gauss requires some symmetry -- so the charge distribution is rather restricted (at least locally uniform, most oftten)