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Homework Help: Electric Field Forces in a Hydrogen Molecule

  1. Oct 14, 2007 #1
    Well, I did the problem. The problem is I KNOW my answer is wrong. Anybody have any idea why? (This is what I've worked out with my friend).

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

    Imagine a sphere of radius a filled with negative charge of uniform density, the total charge being equivalent to that of two electrons. Imbed in this jelly of negative charge two protons and assume that in spite of their presence the negative charge distribution remains uniform. Where must the protons be located so that the force on each of them is zero? (This is a surprisingly realistic caricature of a hydrogen molecule; the magic that keeps the electron cloud in the molecule from collapsing around the protons is explained by quantum mechanics!)

    [Ahhhh logically when I try to think about this, I don't understand how there's ANY way there could be zero force on the protons in a sphere of negative electric charge. But that's non-mathematical.]

    2. Relevant equations

    Gauss's Law (sphere)
    [tex] E = \frac{Q}{r^2} [/tex]
    q = e (charge of an electron)
    a = radius of sphere
    r = radius of Gaussian sphere

    3. The attempt at a solution

    So, supposedly, I'm supposed to take a Gaussian surface/sphere within the sphere on which the charges will rest.

    The charge density of the sphere is
    [tex]\frac{-2q}{\frac{4}{3}\pi a^3}[/tex]
    or [tex]\frac{-3q}{2\pi a^3}[/tex].

    Therefore the charge on my Gaussian sphere is the charge density times the volume of the Gaussian sphere, or
    [tex](\frac{-3q}{2\pi a^3})(\frac{4}{3}\pi r^3)[/tex]

    [tex]E = q_total/r^2[/tex]
    Total charge is 2 electrons - charge inside Gaussian sphere
    [tex]q_{total} = 2q - \frac{2qr^3}{a^3}[/tex]
    [tex]E = \frac{1}{r^2}(2q - \frac{2qr^3}{a^3}[/tex]
    [tex]E = \frac{2q}{r^2} - \frac{2qr}{a^3}[/tex]

    When the E field is equal to zero, there is no force.

    [tex]0 = \frac{2q}{r^2} - \frac{2qr}{a^3}[/tex]
    [tex] \frac{2q}{r^2} = \frac{2qr}{a^3} [/tex]
    [tex]r^3 = a^3[/tex]
    [tex]r = a[/tex]

    The more I look at this the more it doesn't make sense, but I can't pinpoint where the "wrong" is.
  2. jcsd
  3. Oct 14, 2007 #2


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    so far this looks good. seems like you forgot a minus sign at the end.

    I don't understand why you did this. The field is the total "enclosed charge" divided by r^2... why are you taking 2q and subtracting...

    Find the field due to the electrons... taking the above into account.

    Now we know that the field due to the electrons is in the radial direction... so the only way that the protons can be placed in the field motionless, if the field that the proton creates a field in the radial direction...

    how must the protons be placed to cancel the field due to the electrons?
  4. Oct 14, 2007 #3
    correction: So is the field due to the electrons with just [tex]E = (-\frac{2qr^3}{a^2})(\frac{1}{r^2}) = -\frac{2qr}{a^3}[/tex]

    I really have no idea how the protons can cancel the electrons unless they were at the same point? Is there any other hint you could give? Thanks again :(
    Last edited: Oct 14, 2007
  5. Oct 14, 2007 #4


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    Since the field due to the electrons acts radially... the field due to the protons needs to act radially too...

    The protons have to be located opposite each other. ie both protons are located at a radius r from the center opposite each other...

    So what's the field at one of the protons? You've got the field due to the electrons + the field due to the other proton. The 2 need to add to zero...
  6. Oct 15, 2007 #5
    Hmm, I think I am starting to see it. Since the proton will be on the Gaussian sphere at radius r away, for a proton [tex]E = \frac{q}{r^2}[/tex].

    [tex]E = -\frac{2qr}{a^3} + \frac{q}{r^2} = 0[/tex]?
  7. Oct 15, 2007 #6


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    yes, except the field due to the proton will be:

    [tex]E = \frac{q}{(2r)^2}[/tex]

    so set:

    [tex]E = -\frac{2qr}{a^3} + \frac{q}{(2r)^2} = 0[/tex][/QUOTE]
  8. Oct 15, 2007 #7
    Ah, 2r since it's a diameter away. Thank you so, so much. I think I understand this now. Yay for starting to get E&M :).
  9. Oct 15, 2007 #8


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    :) no prob.
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