Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Spherical Electric Field - displacing a nucleuo

  1. Sep 13, 2006 #1
    Spherical Electric Field - displacing a nucleus

    I'm stuck on this question.

    Q) A simple classical model of an atom consists of a nucleus a postive charge [itex] N\,e [/itex] surrounded by a spherical electron cloud of hte same total negative charge. ([itex] N [/itex] is the atomic number and [itex] e [/itex] is the magnitude of the electric charge) An external electric field [itex] E_0 [/itex] will cause the nucleus to be displaced a distance [itex] r_0 [/itex] from the center of the electron cloud, thus polarizing the atom. Assuming a uniform charge distribution within the electron cloud of radius [itex] b [/itex], find [/itex] r_0 [/itex].

    A) Well I have the solution... but I don't understand it.

    This is what the book does.

    First find the electric field of the electron cloud.
    [tex] \epsilon_0 = \oint_S \vec E_{el} \cdot d\vec s = 4 \pi R^2 \epsilon_0 E_R(R) = Q_{enc} [/tex]

    [tex] \vec E_{el} = \hat R \frac{Q_{enc}}{4 \pi \epsilon_0 R^2} [/tex]

    So this vector function gives the electric field of the electron cloud as a function of distance from the orgin, correct?

    Now the book moves onto saying that there are two distinct regions to consider:
    (1) [tex] 0 \leq R \leq b [/tex]
    (2) [tex] r \geq b [/tex]

    It says,
    However since [tex] r_0 > b [/tex] implies ionization of the atom, we will only determine [tex] E_{el} [/tex] for [itex] R < b [/itex].

    I don't understand this. If we chose a point [itex] P > b [/itex] how does this imply ionization of the atom.
    Wouldn't chosing a point just mean,
    [tex] \vec E_{el} = \hat R \frac{Q_{enc}}{4 \pi \epsilon^2 (R_P)^2} [/tex]
    where [itex] R_P [/itex] is the distance from the orgin to P.

    doesn't this say, what is the electric field at this point. How does this have anything to do with removing the proton?

    Would someone explain this?

    The book calculates [itex] Q_{enc} [/itex] as follows:

    The charge density of the electron cloud is given by:
    [tex] \rho = \frac{-N\,e}{4/3 \, \pi b^3} [/tex]
    For [itex] 0 \leq R \leq b [/itex] we have:

    [tex] Q_{enc} = \rho \frac{4}{3} \pi R^3 = -N\,e \frac{R^3}{b^3} [/tex]

    Why doesn't the [itex] Q_{enc} [/itex] enclose the nucleus?

    I really just need some hand holding on this, because I'm lost and I have spent way to much time trying to figure this out. Thanks in advance!
    Last edited: Sep 13, 2006
  2. jcsd
  3. Sep 13, 2006 #2
    The condition r_0>b means that the separation of the nucleus and the electron cloud is greater than the radius of the electron cloud, which means the nucleus is outside of the electron cloud. This is not considered as they'd repel, and you wouldn't have a polarized atom.

    For the second thing you asked about, remember that Q_enc is only the charge of the electron cloud. It would be zero otherwise! You need to treat the field from the electrons and the nucleus separately because of the difference between the centre of the cloud and the nucleus. You have to add the fields from the electrons and the nucleus for the total field, remembering you'd have Q_nuc/(4pi e0 (R+r0)^2) for the nucleus.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook