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Calculate electric field produced by total charge density

  1. Sep 19, 2013 #1
    Problem:

    In a neutral He atom consisting of a positively charged nucleus of charge ##2q## and two electrons each of charge ##-q##, the volume charge density for the nucleus and for the single electron cloud are respectively given by [tex]\rho_n(\vec{r})=2q\delta^3(\vec{r})[/tex] and [tex] \rho_e(\vec{r})=-\frac{8q}{\pi {a_0}^3}e^{-4r/{a_0}}[/tex]

    Using Gauss's law, calculate the electric field ##\vec{E}## produced by the total charge density [tex]\rho(\vec{r})=\rho_n(\vec{r}) + \rho_e(\vec{r})[/tex].

    Attempt at a Solution:

    I'm sure I'll be required to use the integral form of Gauss's law, namely [itex]\oint \vec{E} \cdot d\vec{A}=\frac{Q_{enc}}{\epsilon_0}[/itex].

    It's easy to show that the total charge of the electron cloud is equal to ##-q##. Thus, since the nucleus contributes ##2q## and we have, ##Q_{enc}=2q-q=q##; that is, if we have construct a spherical Gaussian surface enclosing the nucleus and one of the electrons. Is this the correct approach?
     
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  3. Sep 19, 2013 #2

    TSny

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    The electron cloud of one of the electrons extends all the way to infinity. So, if your Gaussian surface has a finite radius, it will not enclose all of the charge of the electron.

    So, think about how you can find the charge inside the Gaussian surface.
     
  4. Sep 19, 2013 #3
    Wouldn't the only way to enclose all of the charge of the electron be to have a sphere with an infinite radius act as our Gaussian surface?
     
  5. Sep 19, 2013 #4

    TSny

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    Why do you want to enclose all of the charge? To find the electric field at a distance ##r## from the nucleus, you'll need a Gaussian surface of radius ##r##. So, you're going to have to find the electric charge enclosed inside that Gaussian surface.
     
  6. Sep 19, 2013 #5
    Oh right! So I'll wind up with an equation for the electric field ##\vec{E}## as a function of ##r##.

    So I start with the LHS of Gauss's law: ##\oint \vec{E} \cdot d\vec{A}##

    ##\vec{E}## and ##d\vec{A}## will be parallel, so we get ##\int |\vec{E}|d\vec{A}##. And since the magnitude of ##\vec{E}## will be constant over the surface, we have ##|\vec{E}| \int d\vec{a}=|\vec{E}|4\pi r^2##.

    Comparing to the RHS:

    ##|\vec{E}|4\pi r^2=\frac{Q_{enc}}{\epsilon_0}##

    So wouldn't the enclosed charge be ##2q## (due to the nucleus) plus whatever fraction of the electron cloud is enclosed in the Gaussian surface?

    EDIT: Left out ##|\vec{E}|##
     
    Last edited: Sep 19, 2013
  7. Sep 19, 2013 #6

    TSny

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    Yes, that's right. (I think you left out E in the last equation.)
     
  8. Sep 19, 2013 #7
    Okay thanks!

    So I'm getting ##\vec{E}=\frac{1}{4\pi \epsilon_0}\frac{q}{r^2}\hat{r}## as my final answer.

    (Fixed the error in my previous post as well.)

    EDIT: Oh wait... pretty sure this is wrong because I didn't account for the electron cloud only partially contributing to the enclosed charge. So should it be ##Q_{enc}=2q+\int_0^r \rho_e(\vec{r}) dr##?
     
  9. Sep 19, 2013 #8

    TSny

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    No, that's not the right answer.

    How did you find ##Q_{enc}## (the net charge inside the Guassian surface of radius r)?
     
  10. Sep 19, 2013 #9
    I don't know if you saw the edit to my last post, but I realized it was wrong right after I posted it. But the charge enclosed in the Gaussian surface should be ##Q_{enc}=2q+\int_0^r \rho_e(\vec{r}) dr##, I believe.
     
  11. Sep 19, 2013 #10

    TSny

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    Almost. Note that ##\rho_e(r)## is a volume charge density. So, you should integrate over the volume enclosed by the Gaussian surface.
     
  12. Sep 19, 2013 #11
    Darn, so close.

    So then I have ##Q_{enc}=2q+\int_0^r \rho_e(\vec{r})d\tau##.

    And the integral is then ##\int_0^r \rho_e(\vec{r})d\tau=\int_0^{2 \pi} \int_0^{\pi} \int_0^{r}(-\frac{8q}{\pi {a_0}^3}e^{-4r/{a_0}})r^2sin\phi \ dr \ d\theta \ d\phi##.
     
  13. Sep 19, 2013 #12

    TSny

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    Yes, that's it.
     
  14. Sep 19, 2013 #13
    One last thing:

    ##2q## comes from the fact that [tex]\int_0^r \rho_n (\vec{r}) d\tau=\int \int \int 2q \delta^3(\vec{r})r^2 sin\phi \ dr \ d\theta \ d\phi=2q \int \int \int \delta^3(\vec{r})r^2 sin\phi \ dr \ d\theta \ d\phi = 2q \cdot 1 = 2q[/tex] Correct?
     
  15. Sep 19, 2013 #14

    TSny

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    Essentially, yes. Working with the the Dirac delta function in spherical coordinates is a little tricky.

    See for example equation 51 here: http://mathworld.wolfram.com/DeltaFunction.html Note that in spherical coordinates your integration over r starts at r = 0, so the integration region doesn't really "contain" r = 0 (the nucleus) as an interior point in the region of integration. That makes me a little nervous.

    To me, it's easier to think of going to Cartesian coordinates for taking care of the delta function at the origin

    ##\delta^3(\vec{r}) = \delta(x)\delta(y)\delta(z)##.

    But stick with spherical coordinates for the integration of the electron charge distribution.
     
  16. Sep 20, 2013 #15

    vanhees71

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    That's correct. Spherical coordinates are singular at [itex]\vec{r}=0[/itex] (even along the entire polar axis), and thus you cannot express properly [itex]\delta^{(3)}(\vec{r})[/itex] in terms of spherical coordinates.
     
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