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I just thought of this problem. In order to use Gauss' Law, we need a closed surface. Can any such imaginary closed surface be constructed in this case? In short, will any flux be coming out of the closed end?

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- Thread starter Wrichik Basu
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Why would this matter? You wanted the flux that goes through the other end. The only thing that matters is how far away that surface is, and its surface...

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I just thought of this problem. In order to use Gauss' Law, we need a closed surface. Can any such imaginary closed surface be constructed in this case? In short, will any flux be coming out of the closed end?

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Any closed surface is a Gaussian surface, so: yes!Wrichik Basu said:Can any such imaginary closed surface be constructed in this case?

" In short " meaning "the same question in other words"?Wrichik Basu said:In short, will any flux be coming out of the closed end?

As a completely diffferent question: yes.

PLease share your thoughts, just like if this were a homework exercise ...

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You know the electric field from a point charge, so just integrate that over the closed end to get the flux.Wrichik Basu said:How much should be the flux coming out from the closed end?

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Let me clear it up a bit. Sorry for the confusing terminology.BvU said:Hollow conducting cylinder ? Height/diameter >> 1 ? Not grounded, I presume ?

Any closed surface is a Gaussian surface, so: yes!

" In short " meaning "the same question in other words"?

As a completely diffferent question: yes.

PLease share your thoughts, just like if this were a homework exercise ...

Yes, it is a conducting cylinder. Both height and diameter are >> 1, and it is not grounded.

What I wanted to say was, since the cylinder had one side open, could I use Gauss' Law to compute the electric flux on the closed end? Gauss' Law requires a closed surface to be constructed in order to find the flux. How do I construct the closed surface so that I can compute the flux on the closed end of the cylinder?

But @Dale has given a solution of not going to Gauss' Law in this case, and thanks for that.

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Could you elaborate ?Dale said:You know the electric field from a point charge, so just integrate that over the closed end to get the flux.

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The flux is ##\Phi=\int\int_S E\cdot dS##. Since you know E (Coulombs law) and S (end of cylinder) you can just directly calculate ##\Phi## without using Gauss’ lawBvU said:Could you elaborate ?

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How does one find ##\vec E\;## for such a configuration ?

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Wrichik Basu said:

I just thought of this problem. In order to use Gauss' Law, we need a closed surface. Can any such imaginary closed surface be constructed in this case? In short, will any flux be coming out of the closed end?

This appears to be nothing more than the problem of a point charged enclosed in a cylindrical Gaussian surface, is it not? All you are interested in is the electric flux passing through the surface at one end of the cylinder.

Zz.

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Coulomb’s lawBvU said:How does one find ##\vec E\;## for such a configuration ?

Edit: oops, I see now that the cylinder is not just a partial Gaussian surface, but an actual conductor.

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That is what we call the problem statement in the homework fora. 'nothing more' and 'enclosed' appears misplaced to me. I am curious to find a handle to work out a solution. Does not seem trivial at all to meZapperZ said:This appears to be nothing more than the problem of a point charge enclosed in a cylindrical Gaussian surface, is it not? All you are interested in is the electric flux passing through the surface at one end of the cylinder.

Zz.

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Thanks for the diagram, I meant that.BvU said:That is what we call the problem statement in the homework fora. 'nothing more' and 'enclosed' appears misplaced to me. I am curious to find a handle to work out a solution. Does not seem trivial at all to me

View attachment 227263

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As per the diagram given by @BvU, one side of the cylinder is open. That's where the problem lies.ZapperZ said:This appears to be nothing more than the problem of a point charged enclosed in a cylindrical Gaussian surface, is it not? All you are interested in is the electric flux passing through the surface at one end of the cylinder.

Zz.

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Wrichik Basu said:As per the diagram given by @BvU, one side of the cylinder is open. That's where the problem lies.

Why would this matter? You wanted the flux that goes through the other end. The only thing that matters is how far away that surface is, and its surface area.

Zz.

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It would be distorted, because in the case where there isn’t a can the E field is purely radial, but in the case where there is a can the field must be perpendicular to the surface of the can.BvU said:

If it were a flat box then I would use the method of images, but in this case I would need to numerically solve the electrostatic equations. There might be a more elegant method, but I don’t know it.

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