Mean Value Theorem of Electrostatics

[pmb_phy]

In "Classical Electrodynamics - 3rd Ed.," J.D. Jackson has an exercise, 1.10, to derive the mean value theorem of electrostatics. Does anyone know of a derivation which is located on the web?

Pete

 

[dextercioby]

Yes, this one

"The average value of the potential over the spherical surface is

\bar{\Phi} =\frac{1}{4\pi R}\int \Phi \ dA

If u imagine the surface of the sphere as discretized, u can rewrite the integral as an infinite sum

\frac{1}{a}\int \ dA =\Sigma_{area}

Then, takin the derivative wrt to R

\frac{d\bar{\Phi}}{dR}=\frac{d}{dR}\Sigma \Phi=\Sigma \frac{d\Phi}{dR}

Convert the infinite sum back to an integral & get

\frac{d\bar{\Phi}}{dR}=\frac{1}{4\pi R^{2}} \int \frac{d\Phi}{dR} \ dA

Now, using that

\frac{d\Phi}{dR}=-E and Gauss's law (no charge inside the sphere) that gives

\frac{d\bar{Phi}}{dR}=0 \Rightarrow \bar{\Phi}_{surface}=\Phi_{center}

Q.E.D.


Daniel.

 

[pmb_phy]

"The average value of the potential over the spherical surface is
\bar{\Phi} =\frac{1}{4\pi R}\int \Phi \ dA

No. It isn't. Its
\bar{\Phi} =\frac{1}{A}\int d\Phi
where A is the area of the sphere.

Pete

 

[qbert]

This isn't quite right either.

No. It isn't. Its
\bar{\Phi} =\frac{1}{A}\int d\Phi
where A is the area of the sphere.
Pete

It isn't clear what you mean by \int d\Phi
since usually the integral of an exact differential
is just the function evaluated at the endpoints.

Or alternatively
\int d\Phi = \int \frac{\partial \Phi}{\partial x} dx +<br /> \int \frac{\partial \Phi}{\partial y} dy +<br /> \int \frac{\partial \Phi}{\partial z} dz
which is, of course, a line integral (which isn't what we
want).

\bar{\Phi} =\frac{1}{4\pi R}\int \Phi \ dA

This doesn't have the right dimensions. On the left
we have dimensions of [Phi]; and on the right we have
dimensions of [Phi]x[Area]/[Length].

I propose another formula:
<br /> \bar{\Phi} = \frac{1}{4 \pi R^2} \int_A \Phi dA <br />

This also has the advantage that if Phi is constant
on the sphere we have
<br /> \bar{\Phi} = \frac{1}{4 \pi R^2} \int_A \Phi dA = <br /> \frac{1}{4 \pi R^2} \Phi \int_A dA = <br /> \frac{1}{4 \pi R^2} \Phi ( 4 \pi R^2 ) =<br /> \Phi<br />

 

[pmb_phy]

This isn't quite right either.
It isn't clear what you mean by \int d\Phi
since usually the integral of an exact differential
is just the function evaluated at the endpoints.

That would be true if you were talking about a line integral rather than in this case where the integral is a surface integral. What d\Phi is? Hmmm. Perhaps you're right. I can't say that is a small element of \Phi since that makes no sense to me at the moment. Thanks.

Pete

 

[robphy]

I propose another formula:
<br /> \bar{\Phi} = \frac{1}{4 \pi R^2} \int_A \Phi dA <br />

May I propose a more [notationally] precise formula:
<br /> \bar{\Phi} = \frac{1}{A} \int_A \Phi dA <br />
or a more [conceptually] precise formula for this average:
<br /> \bar{\Phi} = \frac{ \int_A \Phi dA } {\int_A dA} <br />

 

[qbert]

Ah, pedantry, thou most noble virtue. You are, of course, correct. Now the next person (who reads this thread) who needs a 2-D average will know the correct generalization.

 

[dextercioby]

I admit, it was R^2 in the denominator...

Daniel.

 

[robphy]

Ah, pedantry, thou most noble virtue. You are, of course, correct. Now the next person (who reads this thread) who needs a 2-D average will know the correct generalization.

This is a Homework subforum... and, in my experience, my students appreciate seeing the general idea in a consistent memorable notation rather than a special case. After all, there was some confusion with this formula.

Although the notation "A" [implicitly] suggests an area average, it is certainly not necessarily so.

 

[pmb_phy]

This is a Homework subforum... and, in my experience, my students appreciate seeing the general idea in a consistent memorable notation rather than a special case. After all, there was some confusion with this formula.
Although the notation "A" [implicitly] suggests an area average, it is certainly not necessarily so.

Hi Rob

Do you know where I can find a solution? Jackson seems to want the student to use Green's theorm to solve this.

Pete

 

[alerodri115]

Hope this link helps

http://faculty.cua.edu/sober/536/meanvalue.pdf

 

[jdwood983]

You do realize that this post is coming onto 4 years old? That the original poster hasn't asked a question for well over a year?

There is usually no reason to resurrect threads that are more than 2 months old, let alone ones that are 4 years old.