Understanding Gauss's Law in Cylindrical Shells of Non-Infinite Length

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Gauss's Law indicates that inside an infinitely long cylindrical shell, the electric field is zero due to symmetry. However, for a cylindrical shell of finite length, the electric field is not zero inside because the ends of the shell disrupt the symmetry. When a positively charged washer is introduced near an open-ended cylindrical shell, the electric field will vary within the shell, with the strongest fields likely near the ends. Understanding why the field is non-zero in finite-length tubes involves examining the breakdown of symmetry compared to infinite tubes. This foundational knowledge aids in analyzing more complex scenarios involving cylindrical shells.
PhDnotForMe
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My question is going to be rather specific. I am trying to understand how Gauss's law applies to this scenario. I know if a cylindrical shell is infinitely long, and there is an external electric field, the inside of the shell will have an electric field of zero everywhere. I am wondering what happens when the shell is not infinitely long. I would assume the general answer would be no, the electric field inside the cylindrical shell would not be zero.

But what if we introduce symmetry. Say we have a cylindrical open shell with h=8 units and r=0.5 units and we put it through the hole of a washer so that each endpoint of the shell is 4 units away from the washer. We give the washer a positive charge. Will the inside of the cylindrical shell be zero everywhere? And if not, why not and which regions of the inside of the shell would have the highest electric field? Thanks.
 
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Before you take on this question, try a simpler one: Why is the field not zero everywhere inside a cylindrical shell of finite length (and open ends - if the ends are closed the field will be zero inside no matter what).
 
Nugatory said:
Before you take on this question, try a simpler one: Why is the field not zero everywhere inside a cylindrical shell of finite length (and open ends - if the ends are closed the field will be zero inside no matter what).
My answer to this would be because it is not infinite or closed. What do you think is the answer?
 
PhDnotForMe said:
My answer to this would be because it is not infinite or closed.
Yes, but why is it that that way? Why is it that the field is non-zero inside a tube of finite length but zero inside a tube of infinite length? That's an easier question to answer: you can look at the proof that the field is zero for the infinite tube; identify the point at which it depends on the infinite length; and imagine how the field will behave without that assumption.

And once you've gone through that exercise, you'll be able to see for yourself what's going on in the more complex problem in your orignal post.
 
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