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Q: Field in and around a conducting hemispherical shell

Aayush SethiaSay I have a hemispherical shell and I want to find the electric field inside the shell, assuming that it does not create much of a difference even if I consider that the shell is closed from the bottom or not. Now I want to find whether the electric field inside the shell is zero or not. The re...

You haven’t specified whether the shell is conducting or not. And, if not, you haven’t specified whether the surface charge density on it is uniform.
Have you learned how the field at a charged surface is related to the charge density?
it is conducting , yup it is sigma over epilson not for field
it is conducting This information belongs in the question, not in a comment. You can edit your question.
I know the field at the centre of a hemispherical shell. How do you know that? You don’t know how the charge is distributed.
Could I say that the value of field which is inside the centre of the shell is also same as the value at the surface of the shell? Why do you think they would be the same?
Finding the potential and/or field of a conducting hemispherical shell is a difficult problem. See this question, my answer, and the references within.
Sir, I assuming uniform charge distibution at the surface of the hemisphere
Sir , I actually want to draw an analogy between and a metallic solid hemisphere and a holow metallic hemisphere , as both are conducting so could I say that as inside a solid metallic hemisphere the field is zero then inside a hollow one also the filed should be zero ?
I assuming uniform charge distibution at the surface of the hemisphere. You said “it is conducting”. A conducting hemispherical shell will not have a uniform charge distribution. You cannot make both assumptions.
Could I say that as inside a solid metallic hemisphere the field is zero then inside a hollow one also the filed should be zero? No, because solid metal and empty space are very different things.
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Sir please see the latest edit as well
Sir may I know why I cannot make both assumptions together ?
As I already explained, a conducting hemispherical shell does not have a uniform charge distribution. Look at the pictures in the answer I linked to. Do you see how nonuniform it is?
So tell me sir if I have shell and I want to find field at centre then is it not sigma / 4epsilon not ?
where sigma is charge density
byjus.com/question-answer/… are they wrong here ?
It makes no sense to talk about “sigma” as a constant if the hemisphere is conducting. The charge density depends on position in that case.
No, they’re not wrong. They’re not considering a conducting hemisphere.
I think you are confused about what kind of hemisphere you are considering.
Sir then if it is non conducting then can I say that the field inside is 0 ?
No, you cannot. The field inside is not zero.
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Ok sir and what is the problem with the gaussian surface proof ?
Sir at last please tell me the problem with the gaussian surface proof ?
The proof that I have attached with the post ?
The system does not have enough symmetry for you to be able to apply Gauss’ Law.
Your title still says that the hemisphere is conducting, but apparently you don’t mean this. This discussion is too long and I won’t continue it.
I wanted to hear the symmetry argument thank you very much

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