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Homework Help: A hydrogen atom in uniform electric field

  1. Dec 30, 2005 #1
    Assume there's a hydrogen atom. Its proton(charge +e) is at the center of a spherical electron cloud(-e) which radius is 10Å and the electrons are uniformly distributed at the sphere's surface. If we now put it into a uniform electric field [tex]E_0[/tex], what is the distance [tex] \delta [/tex] between the proton and the center of the spherical electric cloud.

    This a problem of an college enterance exam(physics department) and I'm only a high school student. I have no idea about this. Moreover, I thought if we put the atom into a electric field, the electric cloud will be distorted and will no longer be a spherical. But why the problem ask the distance between the proton and the center of the spherical

    Plz Help Me! thank you very much!
    Last edited: Dec 31, 2005
  2. jcsd
  3. Dec 30, 2005 #2

    Andrew Mason

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    All you have to know is Coulomb's law.

    The electric field gives rise to two forces. One on the centre of the electron cloud and one on the proton. What is the expression for that force in terms of the charge and electric field?

    Since the charges are of opposite sign, what does that tell you about the direction of the forces? What does that do to the distance between the proton and the centre of the electron cloud?

    Then think about the force between the proton and the electron cloud that tends to pull them back together (ie. opposes the forces of the electric field). At some distance, the force pulling them apart is equal to the force pulling them together.

    So work out the expression for those forces, equate them and solve for that distance.

  4. Dec 30, 2005 #3
    I have some problems. If we put the atom into the field and the electrons still distributed uniformely, how we calculate the force between electrons cloud and proton? If the proton is inside the spherical, isn't the force between them be zero?
    And if electron cloud been distorted, it seems more complex!
    Last edited: Dec 30, 2005
  5. Dec 30, 2005 #4
    The force on proton due to field is
    [tex] F_p=eE_0 [/tex]
    and force on electrons due to field is
    [tex] F_e=eE_0 [/tex]
    These two force tends to pull proton and electrons apart.
    And I think the force between them try to pull them together, which is the force I don't know how to deal with.
  6. Dec 30, 2005 #5

    Andrew Mason

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    That is the coulomb force:

    [tex]F = \frac{q_1q_2}{4\pi\epsilon_0 r^2}[/tex]

  7. Dec 30, 2005 #6
    If the electron sphere is uniform, the forceinside the sphere is zero. Just like the graviational force inside a hollow planet is zero.
    So how we use [tex]\frac{q_1 q_2}{4\pi\epsilon_0 r^2}[/tex] ?
  8. Dec 30, 2005 #7


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    Don't we model the force on the surface as a point charge in the centre of the sphere?
  9. Dec 31, 2005 #8
    If the proton is out fo the sphere, of course we can see the electron cloud as a point at the sphere's center. But the point is, even we put the atom into a electric field, the proton doesn't run out of the electron cloud, it just shifts, so the proton and the center of the electron cloud sphere won't coincide. So the proton is still in the electron cloud sphere. Hence the whole force on proton due to electron cloud is zero. (I think)
  10. Dec 31, 2005 #9


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    The electric force that a test charge experiences while completely inside a uniformly charged spherical shell is zero. However, the force such a charge feels while on the surface of a uniformly charged sphere is equal to the force it would feel if all of the charge of the sphere were concentrated at the center. The situation we have here is different. We have a charge inside a ball of uniformly distributed charge. Think of this as a combination of the previous two scenerios.
  11. Dec 31, 2005 #10
    Sorry, I frogot to translate a important imformation of the problem. The problem said:There's a spherical electron cloud 10Å from the centre, charged -e, and charge is uniformly distributed on the spherical shell.
    Sorry for making mistake.
  12. Jan 1, 2006 #11
    Is the shell theory(a charge in a hollow and charge uniformly distrubuted sphere experiences no force due to the sphere) based on that the test charge won't change the charge distribition of the sphere? If it is, in this case, the charge of electron cloud sphere won't distributed uniformly, since there charge have the same magnitude e. SO HOW WE SOLVE THIS PROBLEM??
    Last edited: Jan 1, 2006
  13. Jan 1, 2006 #12


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    The whole problem is an unrealistic scenerio. You never have a uniformly distributed electron cloud in a perfectly thin sphere around the nucleus, and if there was an electric field it would distort the shape of an electron cloud rather than simply translating it. However, the problem says that the cloud is sherical even after application of the E field. It says the charge is uniformly distributed on the shere before the application of the field, which is really not too helpful. I think its safe to assume that they mean the charge distribution remains uniform because there would be no way of solving the problem otherwise because if you did find the shape the cloud assumed it would not be spherical in the first place, so it would be pretty impossible to find the charge distribution on the spherical electron cloud. It is poorly worded.
  14. Jan 2, 2006 #13
    Well... I agree with you.
    Last edited: Jan 2, 2006
  15. Jan 2, 2006 #14
    i believe this problem is discussed in griffiths's E&M book, chapter 4. you need to know the field at a distance from the center inside a spherical charge distribution, and find the point where that field counterbalances the external field. the displacements are small, so you can assume the charge distribution remains uniform and spherical.

    hmm....nevermind, i just reread the OP and saw that it was for a spherical SHELL...
    Last edited: Jan 2, 2006
  16. Jan 2, 2006 #15
    So it is possible to solve this problem if the electron cloud is a spherical ball which charges are uniformly distributed inside it?
  17. Jan 2, 2006 #16
    yes...by gauss's law, the field inside such a distribution increases linearly with distance from the center; thus, the point at which the field is equal to the external field and opposite in direction will be an equilibrium point for the positive charge.
  18. Jan 3, 2006 #17
    I get it. thanks a lot!
  19. Oct 11, 2009 #18
    distance = 4 X pi X Eo X R^3 X E /e

    Eo= petmativity of the free space
    E= the external electric field
    e= charge
    R= atomic distance
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