Homework Help: Finding E and V of an atom

1. Sep 25, 2010

xxbigelxx

1. The problem statement, all variables and given/known data

The electrons of the uranium atom can be approximated as a uniform density sphere with radius 7.4 x 10-13 cm. The nucleus is a point at the center with a charge of +92e, and there are 92 electrons.
a. Calculate the electric potential and electric field at the surface taking only the electrons into account.
b. Calculate the electric potential and electric field at the surface taking only the nucleus into account.
c. Compare the 2 sets of values. d. A more classical picture has the electrons in planetary orbits. At any instant, there are 92 electrons arrayed uniformly about the sphere. Discuss how this changes the answers to the 3 previous parts.

2. Relevant equations

3. The attempt at a solution

I am not entirely understanding the setup for this problem. I am having a hard time understanding what is being described. If I am understanding it correctly, I would guess I would have to use Gauss Law to get the value for E, and then use that value to find V. Can anyone comment on my thought process?

2. Sep 25, 2010

diazona

That's one way to go about it. Try it and see if you can get it to work.

3. Sep 25, 2010

xxbigelxx

Well wouldn't the values for E be:
due to electrons only.....(-92e/epsilon-naught)(4Pi*r^2) where r= 7.4 x 10-13 cm?
due to nucleus only... the same as above, except it would be positive?

4. Sep 26, 2010

xxbigelxx

I also don't know how to find the potential (and I am not sure if my above post was correct either).

5. Sep 26, 2010

diazona

OK, well your method for the electric field sounds right. What do you know that might allow you to calculate the potential?

6. Sep 26, 2010

xxbigelxx

I was thinking about gradient of E is equal to V?

7. Sep 26, 2010

diazona

It's the other way around,
$$\vec{\nabla}V = \vec{E}$$

8. Sep 26, 2010

xxbigelxx

Oh right that's what I meant (I was typing a bit too quickly). I thought about this technique but I wasn't sure how it would work. I tried a few things on scratch paper, but I kept hitting dead ends.

9. Sep 26, 2010

diazona

The equation as given allows you to calculate E from V by taking the derivative, but you need to go the other way around: you need to calculate V from E. How would you rearrange the equation to do that?

10. Sep 26, 2010

xxbigelxx

Would you have to integrate?

11. Sep 26, 2010

diazona

Yep. Can you write out the formula in full?
$$V(\vec{r}) = \int_{?}^{?}\cdots$$

12. Sep 27, 2010

xxbigelxx

Isn't it just the integral of E.dl from where you are measuring it to where the reference point is?

13. Sep 27, 2010

diazona

Yep, that's correct. How would you apply that here?

14. Sep 27, 2010

xxbigelxx

That's where I'm getting stuck. Isn't the value of E going to be a constant at the edge of the atom? Would it just be that constant value of E times the Radius (dl)?

15. Sep 27, 2010

diazona

Let's back up a bit: where is your reference point? And where are you measuring the field? I think you need to first be sure you know where the two endpoints of the integral are.

(By the way, the integral goes from the reference point to the location where you're measuring the potential, not the other way around.)

16. Sep 27, 2010

xxbigelxx

Would the reference point be at infinity, since the potential there is 0? And we are finding the field at the given value of 'r.' I hope this is right.

17. Sep 27, 2010

diazona

That's right. So you need to integrate the electric field from infinity to the electron sphere's radius, which I'll denote re for simplicity (rather than writing out the number every time).
$$\int_\infty^{r_e} \vec{E}(\vec{r})\cdot\mathrm{d}\vec{l}$$
Can you do something with that?

18. Sep 27, 2010

xxbigelxx

Well I believe that E is constant, so that can be pulled to the front of the integral. That leaves you with E(l1-l2) where l1=r=7.4 x 10-13 cm. What about that infinity? That seems to change things...

19. Sep 27, 2010

diazona

Well, that's incorrect. E is not constant. Why did you think it was?

Can you write a formula for E as a function of position?

20. Sep 27, 2010

xxbigelxx

I thought it was since earlier I said that E was equal to (-92e/epsilon-naught)(4Pi*r^2) where r= 7.4 x 10-13 cm. I thought all of those values were constant at the surface of the sphere, and E was constant as a result.

21. Sep 27, 2010

diazona

Ah, well you found the value of E at the surface of the sphere - that is, you found it at one particular location. That doesn't mean it's constant. In order for E to be constant, it would have to have that value at all locations in space.

Do you understand that in order to do the integral, you need to have values for E not only at the surface of the sphere, but all the way out to infinity?

22. Sep 27, 2010

xxbigelxx

Ohh I think I understand now. So my integral would be Edl where E=-92e/epsilon-naught)(4Pi*r^2) and dl=dr? Also the bounds for r are from infinity to the given value of r?

23. Sep 27, 2010

diazona

Yeah, that's more like it.

24. Sep 28, 2010

xxbigelxx

Ok so I got V=-92e/[(1.5E-35)*epsilon-naught]

25. Sep 28, 2010

diazona

OK, well if you did the math correctly, I suppose that would be the answer. (I haven't calculated it myself)

Although: make sure you include the correct units, and think about whether the sign of the result (positive or negative) makes sense.