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:
$$\bar{\Phi} = \frac{1}{4 \pi R^2} \int_A \Phi dA$$

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

 

[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:
$$\bar{\Phi} = \frac{1}{4 \pi R^2} \int_A \Phi dA$$

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

 

[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 theorem 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.