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Deriving potential energy of an electron inside a nucleus

  1. Jan 20, 2014 #1
    1. The problem statement, all variables and given/known data

    Derive equation below. It is the potential energy of an electron inside a nucleus assumed to be a uniformly charged sphere of R.

    2. Relevant equations

    V'(r) =( -Ze2/4∏ε0R)(3/2 - (1/2)(r/R)^2)

    3. The attempt at a solution

    So I understand that I need to find the

    ∫ from infinity to a point A (A exists inside the uniformly charged sphere)

    I did this by splitting it between infinity to B(on surface of sphere) and A to B.

    I did get the (-Ze^2/4∏ε0R)(1/2)(r/R)^2, but I just am lost for the 3/2. Any help would be wonderful.

    Thanks!
     
  2. jcsd
  3. Jan 20, 2014 #2

    TSny

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    Hello, mafisco. Welcome to PF.

    Sounds like you have the right idea. Can you show more detail of your calculation for each part of the split?
     
  4. Jan 20, 2014 #3
    Hey, thanks! I'm going to be here for a while I believe, taking on a nuclear physics class and it's been a while since I've done physics.

    Alright, so I know that

    E*dA= (Qenc/(ε0*Volume(4/3∏R3))) * new volume(4/3∏r3)

    for the limit integral of A to B.

    Taking Qenc = Ze2 and dA = 4∏r2,

    I get E = Ze2r/(4∏ε0R3)

    With integration, it becomes E = (Ze2r2)/(8∏ε0R3)

    That's the right hand part of what I need, but the left hand part (which I thought would be easy) appears to not give me the 3/2 I need.

    What I tried :-
    E*dA = Qenc/ε0. dA = 4∏r2, Qenc = Ze2.

    I would end up with E = (Ze2)/(4∏ε0r2) which doesn't really give me the result I need after integration.

    Let me know what I can do! Thanks!
     
    Last edited: Jan 20, 2014
  5. Jan 20, 2014 #4

    TSny

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    You should be doing definite integrals. So, you need to substitute in the limits of integration corresponding to points B and A.

    Likewise for the integration outside the sphere from infinity to B.

    (I actually prefer to switch the order of the limits and make up for it by changing the overall sign of the integrals. That way I'm always integrating from small to larger values of r and my lower limits on the integrals are less than the upper limits. So, inside the sphere, I integrate from A to B (on surface) and then outside I integrate from B to infinity. If I don't do it this way I invariably get some signs wrong. :redface:)
     
  6. Jan 22, 2014 #5
    Wonderful.

    That did the trick my man, thank you very much. It seems as if I had forgotten definite integrals quite sadly.

    Hope to see more people like you around, thanks again!
     
  7. Nov 3, 2016 #6
    I don't get the integration limits! Can you please show how you evaluated when applying the limits?
     
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