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Homework Help: Average Potential Electricity and Magnetism

  1. Sep 30, 2008 #1
    1. The problem statement, all variables and given/known data
    Find the average potential over a spherical surface of radius R due to a point charge q located inside. Show that in general: (EQ 1 below), where Vcenter is the potential at the center due to all external charges and Qenc is the total enclosed charge


    2. Relevant equations
    EQ 1 Vave = Vcenter + (Qenc/4*pi*ε₀*R)


    3. The attempt at a solution
    I am just really confused on where to at least get started. I'll be at the computer for awhile so feel free to ask questions I just want to get it started.
     
  2. jcsd
  3. Sep 30, 2008 #2

    gabbagabbahey

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    Try placing a point charge on the z-axis a distance z'<R from the origin. What is the potential [itex]V(\vec{r})[/itex] at a general point [itex]\vec{r}[/itex] in spherical coordinates? Average this potential over the surface of a sphere of radius R. What do you get?
     
  4. Sep 30, 2008 #3
    1/(4*pi*epsilon nought) q/r
     
  5. Sep 30, 2008 #4

    gabbagabbahey

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    Is it really?! I thought the potential due to a point charge located at [itex]\vec{r'}[/itex] was:

    [tex]\frac{1}{4\pi \epsilon_0} \frac{q}{|\vec{r}-\vec{r'}|}[/tex]

    Of course, when the charge is at the origin, [itex]\vec{r'}=0[/itex] and the potential reduces to the one you gave. But(!) what about when the charge is located along the z-axis a distance z' from the origin (i.e.[itex]\vec{r'}=z'\hat{z} \neq 0[/itex]) ??
     
  6. Sep 30, 2008 #5
    not too sure?
     
  7. Sep 30, 2008 #6

    gabbagabbahey

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    Have you learned about vectors yet? If so, what is [itex]|\vec{r}-z'\hat{z}|[/itex] in spherical coordinates?
     
  8. Sep 30, 2008 #7
    i really dont know?
     
  9. Sep 30, 2008 #8

    gabbagabbahey

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    you don't know if you've studied vectors yet?
     
  10. Oct 1, 2008 #9
    I have and I know the potential outside the sphere has to only be in the +z direction because of symmetry, but I dont know the second part of your question.
     
  11. Oct 1, 2008 #10

    gabbagabbahey

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    Well,

    [tex]|\vec{r}-z'\hat{z}|=\sqrt{(r\hat{r}-z'\hat{z}) \cdot (r\hat{r}-z'\hat{z})}=\sqrt{r^2-2rz' (\hat{r} \cdot \hat{z})+z'^2}=\sqrt{r^2-2rz'cos(\theta)+z'^2}[/tex]

    [tex]\Rightarrow V(\vec{r})= \frac{1}{4 \pi \epsilon _0} \frac{q}{\sqrt{r^2-2rz'cos(\theta)+z'^2}}[/tex]

    And so on the spherical surface [itex]r=R[/itex],

    [tex]\Rightarrow V(R,\theta,\phi)= \frac{1}{4 \pi \epsilon _0} \frac{q}{\sqrt{R^2-2Rdcos(\theta)+d^2}}[/tex]

    (where I have set z'=d the distance of the point charge from the origin)

    Do you know how to average a function over a surface?
     
  12. Oct 1, 2008 #11
    no i dont
     
  13. Oct 1, 2008 #12

    gabbagabbahey

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    The average of a function [itex]f(\vec{r})[/itex] over any surface [itex]\mathcal{S}[/itex] is defined as

    [tex]f_{ave}=\frac{\int_{\mathcal{S}} f(\vec{r})da }{\int_\mathcal{S} da}=\frac{1}{A} \int_{\mathcal{S}} f(\vec{r})da[/tex]

    where [itex]A[/itex] is the area of the surface, and [itex]da[/itex] is the infitesimal area element for said surface.

    What is [itex]da[/itex] for a spherical surface of radius R (in spherical coordinates)?

    Can you apply this to [itex]V(R,\theta,\phi)[/itex]?
     
  14. Oct 1, 2008 #13
    da would be r^2sin(theta)drdtheta
    so then
    Vave = 1/a(integral (k q/r)*r^2sin(theta)drdtheta
     
  15. Oct 1, 2008 #14

    gabbagabbahey

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    Does [itex]r[/itex] really vary on a spherical surface of fixed radius [itex]r=R[/itex] ? If not, why is there a [itex]dr[/itex] in your [itex]da[/itex] ? Shouldn't there be a [itex]d\phi[/itex] term instead?

    And why are you using kq/r for [itex]V(R,\theta,\phi)[/itex]?

    What are the limits of integration for [itex]\theta[/itex] and [itex]\phi[/itex]?
     
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