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Help on Electrostatic Force

  1. Aug 28, 2004 #1
    Can someone help me with this problem:

    Consider a simplified model of the singly-ionized helium atom, He+: 2 protons plus 2 neutrons in the nucleus, "orbited" by one electron at a constant distance of 2.65 x 10^-11m.

    a. What is the magnitude of the attractive electrostatic force between the nucleus and electron?

    b. What is the magnitude of the attractive gravitational force between the same two particles?

    I know I have to use Coloumbs Law, but I'm wondering if I have to just plug in numbers. How do I start? How do I get Q1 and Q2?

  2. jcsd
  3. Aug 28, 2004 #2
    I'll give you a clue: the charge on an electon is e = -1.6 x 10 ^-19 C.

    I'll leave you to fogure out what this means for your 2 protons.
  4. Aug 28, 2004 #3
    What do I do with the 2 neutrons?

    I know Q1 = Q2 = -1.6 x 10 ^-19 C.

    So it will look like this:

    F = ((8.988 x 10^9)(1.6 x 10^-19)^2)/2.65 x 10^-11m

    Is the answer for the above equation solve question 'a'?
  5. Aug 28, 2004 #4
    Neutrons have no charge so you don't include them in the EM force, but they have mass so you will have to include them for the gravitational force.0

    And EM foce is an INVERSE SQUARE law, so you should divide over distance squared.
  6. Aug 28, 2004 #5

    Doc Al

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    Staff: Mentor

    In addition, realize that the helium nucleus has two protons.
  7. Aug 28, 2004 #6
    So is the answer to question a, 3.28 x 10^-7?
  8. Aug 29, 2004 #7
    No. If you look above we said 2 protons, and each proton will have a charge of 1.6x10-19
  9. Aug 30, 2004 #8
    Oh, so I have to multiply it by 2? 1.6x10-19 x 2?
    and the answer is for q1 and q2?
  10. Aug 30, 2004 #9
    No just for q1. The force is attractive between 2 protons (q1) in the nucleus and one electron (q2) in orbit. So use Coulombs law for these values.
  11. Aug 31, 2004 #10
    Sorry for asking too many questions,

    so the answer to part 'a' is 6.55 x 10^-7?
  12. Aug 31, 2004 #11
    Thats what I get
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