1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    User Avatar
    Homework Helper
    Gold Member

    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

    User Avatar
    Homework Helper
    Gold Member

    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

    User Avatar
    Homework Helper
    Gold Member

    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

    User Avatar
    Homework Helper
    Gold Member

    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

    User Avatar
    Homework Helper
    Gold Member

    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

    User Avatar
    Homework Helper
    Gold Member

    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

    User Avatar
    Homework Helper
    Gold Member

    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]?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Average Potential Electricity and Magnetism
  1. Magnetic Potential (Replies: 13)

  2. Magnetic Potential (Replies: 4)

Loading...