Average Potential Electricity and Magnetism

[jrc5135]

Homework Statement

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

Homework Equations

EQ 1 Vave = Vcenter + (Qenc/4*pi*ε₀*R)

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.

 

[gabbagabbahey]

Try placing a point charge on the z-axis a distance z'<R from the origin. What is the potential V(\vec{r}) at a general point \vec{r} in spherical coordinates? Average this potential over the surface of a sphere of radius R. What do you get?

 

[jrc5135]

1/(4*pi*epsilon nought) q/r

 

[gabbagabbahey]

Is it really?! I thought the potential due to a point charge located at \vec{r&#039;} was:

\frac{1}{4\pi \epsilon_0} \frac{q}{|\vec{r}-\vec{r&#039;}|}

Of course, when the charge is at the origin, \vec{r&#039;}=0 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.\vec{r&#039;}=z&#039;\hat{z} \neq 0) ??

 

[jrc5135]

not too sure?

 

[gabbagabbahey]

Have you learned about vectors yet? If so, what is |\vec{r}-z&#039;\hat{z}| in spherical coordinates?

 

[jrc5135]

i really don't know?

 

[gabbagabbahey]

you don't know if you've studied vectors yet?

 

[jrc5135]

I have and I know the potential outside the sphere has to only be in the +z direction because of symmetry, but I don't know the second part of your question.

 

[gabbagabbahey]

Well,

|\vec{r}-z&#039;\hat{z}|=\sqrt{(r\hat{r}-z&#039;\hat{z}) \cdot (r\hat{r}-z&#039;\hat{z})}=\sqrt{r^2-2rz&#039; (\hat{r} \cdot \hat{z})+z&#039;^2}=\sqrt{r^2-2rz&#039;cos(\theta)+z&#039;^2}

\Rightarrow V(\vec{r})= \frac{1}{4 \pi \epsilon _0} \frac{q}{\sqrt{r^2-2rz&#039;cos(\theta)+z&#039;^2}}

And so on the spherical surface r=R,

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

(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?

 

[jrc5135]

no i dont

 

[gabbagabbahey]

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

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

where A is the area of the surface, and da is the infitesimal area element for said surface.

What is da for a spherical surface of radius R (in spherical coordinates)?

Can you apply this to V(R,\theta,\phi)?

 

[jrc5135]

da would be r^2sin(theta)drdtheta
so then
Vave = 1/a(integral (k q/r)*r^2sin(theta)drdtheta

 

[gabbagabbahey]

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

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

What are the limits of integration for \theta and \phi?