Infinite long dielectic cylintrical shell of some finite thickness

1. Dec 26, 2005

kant

Imagine a infinitly long cylintrical dielectic shell of thickness n. show that the electric field everywhere inside such shell is zero. What is the electirc field within the thickness of the shell?

2. Dec 26, 2005

marlon

Ok, we got the problem. So, what have you done to solve it ? Show us your progress. We have a certain policy when it comes to Homework Help, that we don't just give away answers. You can read all about this policy in the PF Guidelines.

regards
marlon

EDIT : use Gauss' Law

3. Dec 26, 2005

kant

Hmm.. The problem here is that i don t think one can even implement gauss law. If one s objective is to find the eletric field inside such structure, the best option is perhaps to take the sum of many dE to the point( in the hollow shell) in question. This is not really a H.W question. Is actually a challenge from physics teacher.

4. Dec 26, 2005

marlon

If you have cylindrical symmetry, you can apply Gauss' Law.

marlon

5. Dec 26, 2005

kant

It might work for the outer two region, but not for the region inside the shell.

Last edited: Dec 26, 2005
6. Dec 26, 2005

marlon

Why not ? Prove it.

marlon

7. Dec 26, 2005

kant

Let say that you were to use gauss law within the shell. What do you think the E field will be?

How many sources of E field are there in the problem?

8. Dec 26, 2005

marlon

Please don't turn the question around

You specifically told me that Gauss' Law cannot be applied in this case. Why do you think that ? Prove it.

regards
marlon

9. Dec 26, 2005

kant

Haha... sure.

If i was to do the problem using gauss law. I will have to construct a smell cylindrical shell inside the larger dielectric shell. In this case, the electric field E( x, y) at any point within the dielectric shell is a superposition of the electric field that is 1) due to the charge within our pretend gaussian surface, called it, E ( in) , and 2) the electric field from everywhere else(dielectric shell), called it E( out). E( in ) is obviously zero, because there is no charge within the pretend gaussian surface, but E( out) is not zero.

To proof that E( out) is indeed zero will require coulumbs law.

10. Dec 27, 2005

mukundpa

Electric field due to what?
Is there an external field? Is the dielectric is having an extraneous charge, and if it is, how it is distributed?

Without that how one can think of solving for electric field?

11. Dec 27, 2005

kant

12. Dec 27, 2005

daniel_i_l

You can show that the voltage at any point inside the cylinder is the same as on the surface. If the voltage is always the same then no work is done by the field to move a charge inside the cylinder so the electrical field must be 0.

13. Dec 27, 2005

kant

of course, anyone can convert a dE problem to a dV problem, but the challenge is still there for anyone to show that the E field inside is Zero.

14. Dec 31, 2005

mukundpa

Kant, I still reading it again and again!!!