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Homework Help: Finding E and V of an atom

  1. Sep 25, 2010 #1
    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. jcsd
  3. Sep 25, 2010 #2

    diazona

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    That's one way to go about it. Try it and see if you can get it to work.
     
  4. Sep 25, 2010 #3
    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?
     
  5. Sep 26, 2010 #4
    I also don't know how to find the potential (and I am not sure if my above post was correct either).
     
  6. Sep 26, 2010 #5

    diazona

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    OK, well your method for the electric field sounds right. What do you know that might allow you to calculate the potential?
     
  7. Sep 26, 2010 #6
    I was thinking about gradient of E is equal to V?
     
  8. Sep 26, 2010 #7

    diazona

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    It's the other way around,
    [tex]\vec{\nabla}V = \vec{E}[/tex]
     
  9. Sep 26, 2010 #8
    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.
     
  10. Sep 26, 2010 #9

    diazona

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    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?
     
  11. Sep 26, 2010 #10
    Would you have to integrate?
     
  12. Sep 26, 2010 #11

    diazona

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    Yep. Can you write out the formula in full?
    [tex]V(\vec{r}) = \int_{?}^{?}\cdots[/tex]
     
  13. Sep 27, 2010 #12
    Isn't it just the integral of E.dl from where you are measuring it to where the reference point is?
     
  14. Sep 27, 2010 #13

    diazona

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    Yep, that's correct. How would you apply that here?
     
  15. Sep 27, 2010 #14
    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)?
     
  16. Sep 27, 2010 #15

    diazona

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    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.)
     
  17. Sep 27, 2010 #16
    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.
     
  18. Sep 27, 2010 #17

    diazona

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    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).
    [tex]\int_\infty^{r_e} \vec{E}(\vec{r})\cdot\mathrm{d}\vec{l}[/tex]
    Can you do something with that?
     
  19. Sep 27, 2010 #18
    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...
     
  20. Sep 27, 2010 #19

    diazona

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    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?
     
  21. Sep 27, 2010 #20
    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.
     
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