# Griffiths Problem 4.24

1. Apr 6, 2008

### ehrenfest

1. The problem statement, all variables and given/known data
This question refers to Griffiths E and M book.

An uncharged conducting sphere of radius a is coated with a thick insulating shell (dielectric constant \epsilon_r) out to radius b. This object is now placed in an otherwise uniform electric field $\bfE_0$. Find the electric field in the insulator.

2. Relevant equations

3. The attempt at a solution
So, I wrote down the boundary conditions and I can't seem to find solutions that match them. Since the electric field vanishes inside of a conductor. The solution inside the dielectric must be constant at r=a. Its derivative must vanish at r=a since there is no free charge on that surface. I don't see how we can obtain that unless the potential is constant inside the conductor. But then how can you meet the boundary conditions for the r=b where the potential is not constant!

Last edited: Apr 7, 2008
2. Apr 6, 2008

anyone?

3. Apr 7, 2008

### malawi_glenn

Why dont you show your attempt to solution? You seldom does that..

4. Apr 7, 2008

### ehrenfest

I did actually. I just didn't put it in "The attempt at the solution" section.

EDIT: Now I did.

Last edited: Apr 7, 2008
5. Apr 7, 2008

### faen

I think you're supposed to use the eigenfunction equation for angular momentum (L^2 * psi = h_bar^2 * l(l+1) psi instead of the hamiltionian equation. Then scale the eigenvalue to get the measure in energy. Never done it myself, but thatd be my guess. Remember, the its not charge dependent, the particles could be neutrons.

Last edited: Apr 7, 2008
6. Apr 7, 2008

### Hootenanny

Staff Emeritus
A statement of the question may also be useful...

7. Apr 7, 2008

### ehrenfest

I think you are looking at the wrong book.

8. Apr 7, 2008

### Hootenanny

Staff Emeritus
Why not state the damn problem in your opening post to avoid such confusion?! :grumpy:

9. Apr 7, 2008

### ehrenfest

I wrote this as the first line in my post: "This question refers to Griffiths E and M book." I think the confusion is not my fault.

This is not really something you could help me with unless you have the book (and a lot of people do have the book).

10. Apr 7, 2008

### Hootenanny

Staff Emeritus
That is of course your choice, you are more than welcome to wait for someone who has a copy of that text (with the same edition as you). I was simply offering to help. However, do you not wonder why so many of your threads go unanswered?

Last edited: Apr 7, 2008
11. Apr 7, 2008

### ehrenfest

See the opening post.

12. Apr 7, 2008

### Hootenanny

Staff Emeritus
See pages 97-100 of Electromagnetism by Grant & Philips

13. Apr 7, 2008

### ehrenfest

Can you just tell me what is the flaw in my attempt at the solution?

14. Apr 7, 2008

### genneth

The obvious flaw is that the electric field is not constant at r=a. The lack of free charge implies that the divergence of *D* vanishes.

15. Apr 7, 2008

### ehrenfest

The POTENTIAL must be constant in r = a, correct, since all of the metal sphere is an equipotential.

16. Apr 8, 2008

### siddharth

Not right. Why don't you try writing down what the general form of the solution to Laplace equation would be in the three regions?

This way, you'll see how many constants you need to evaluate, and the corresponding boundary conditions.

Last edited: Apr 8, 2008
17. Apr 8, 2008

### ehrenfest

I don't understand. The potential is constant in the metal sphere. Therefore it must be constant at the boundary.

18. Apr 8, 2008

### siddharth

Constant at the boundary? Do you mean continuous?

What I mean is this.

$$V(r,\theta) = 0, \quad r<a$$
$$V(r,\theta) = ?? \quad a<r<b$$
$$V(r,\theta) = ?? \quad r>b$$

If you can find the functional form $$V(r,\theta)$$ (hint: soln of laplace eqn in spherical coordinates), then you can find the corresponding constants by evaluvating the boundary conditions (ie, continuity at r=a and r=b).

Last edited: Apr 8, 2008
19. Apr 8, 2008

### ehrenfest

What I am saying is that $$V(a,\theta) =\frac{\partial V}{\partial r} \left|_a= 0$$. That is true, right?

That means that the solution inside the solution inside the dielectric cannot have any \theta dependence right or else either the derivative of the potential or the potential itself would have theta dependence at a.

There is NO WAY the electric is not constant at r=a! Then it would not be constant inside the conductor which is absurd.

Hmm. OK I was using the following formula (eqn 4.41) in Griffiths :

$$\epsilon_{above} \vec{E}_{above} \cdot \vec{n} - \epsilon_{below} \vec{E}_{below} \cdot \vec{n} = \rho_f$$

at r=a. I am wondering whether that applies or not because my book says it applies here. Maybe that only applies at the boundary of two dielectrics and a metal is not considered a dielectric i guess?

If that equation does apply at r=a, then genneth must be wrong and the electric field must be constant (actually 0) just above the boundary at r = a.

Last edited: Apr 9, 2008
20. Apr 9, 2008

anyone?