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Magnitude of Electric Field from an Atom

  1. Oct 11, 2011 #1
    1. The problem statement, all variables and given/known data

    The neutral neptunium atom has 93 electrons.
    What is the magnitude of its electric field at a distance of 6.5x10-10 m from the nucleus?

    Hint: The number of protons in a nucleus is equal to the number of electrons in the neutral atom.

    q = 1.6e-19 * 93 = 1.48e-17
    r = 6.5e-10
    k = 8.99e9

    2. Relevant equations

    E = kq / r2

    3. The attempt at a solution

    I would think that it would just be E = kq / r2 since we have all the variables we need.

    E = kq / r2
    E = (8.99e9)(1.48e-17) / (6.5e-10)2
    E = 1.33e-7 / (6.5e-10)2
    E = 3.16e11 => 3.16 x 1011 N/C

    However turns out that isn't the right answer. So i took the hint into consideration, but can't see how it fits in to play. If there are 93 Electrons and 93 Protons wouldn't it mean the charge would just be 0?

    Any help?
     
  2. jcsd
  3. Oct 11, 2011 #2
    Can you apply Gauss's law at this radius?
     
  4. Oct 11, 2011 #3
    Sure, but what will that do?

    Gauss's Law = EA
    = (3.16e11)(4π(6.5e-10)2)
    = 1.67e-6
     
  5. Oct 11, 2011 #4

    SammyS

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    What is the size of such a atom?
     
  6. Oct 11, 2011 #5
    Well, you can relate the charge enclosed within a surface to the electric field at the surface regardless of how the charge is distributed. So at the this radius what charge is enclosed?
     
  7. Oct 11, 2011 #6
    -1.67e-6 = [k(q) / r2] * [4πr2]
    -1.67e-6 / [4πr2] = [k(q) / r2]
    -1.67e-6[r2] / [4πr2] = k(q)
    -1.67e-6 / 4π = k(q)
    -1.33e-7 = k(q)
    q = -1.33e-7 / 8.99e9
    q = -1.48e-17 N/C
     
  8. Oct 11, 2011 #7
  9. Oct 11, 2011 #8

    SammyS

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