Infinite long dielectic cylintrical shell of some finite thickness

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Homework Help Overview

The discussion revolves around an infinitely long cylindrical dielectric shell of finite thickness, focusing on the electric field within and around the shell. Participants are exploring the implications of Gauss' Law and the conditions under which it can be applied to this scenario.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the application of Gauss' Law and question its validity in determining the electric field inside the shell. There are attempts to analyze the electric field contributions from different regions and the implications of charge distribution.

Discussion Status

The discussion is active, with participants questioning assumptions about the applicability of Gauss' Law and the nature of the electric field within the shell. Some guidance has been offered regarding the use of Gauss' Law, but there is no consensus on its effectiveness in this specific case.

Contextual Notes

Participants note potential constraints regarding the charge distribution within the dielectric shell and the need for additional information to fully address the problem. There is also mention of the problem being framed as a challenge rather than a standard homework question.

kant
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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?
 
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kant said:
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?

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
 
marlon said:
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


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

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


It might work for the outer two region, but not for the region inside the shell.
 
Last edited:
kant said:
It might work for the outer two region, but not for the region inside the shell.

Why not ? Prove it.

marlon
 
marlon said:
Why not ? Prove it.
marlon


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?
 
kant said:
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?

Please don't turn the question around:wink:

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

regards
marlon
 
marlon said:
Please don't turn the question around:wink:
You specifically told me that Gauss' Law cannot be applied in this case. Why do you think that ? Prove it.
regards
marlon

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
kant said:
...How many sources of E field are there in the problem?...

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
mukundpa said:
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?


Read it again.
 
  • #12
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
daniel_i_l said:
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.


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
Kant, I still reading it again and again!
 

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