Finding electric flux using Gauss' Law

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Say you have a hollow cylinder, whose one side is open. Now, you pace a positive charge ##Q## at the centre of this open end (such that it is just inside the cylinder). How much should be the flux coming out from the closed end?

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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  • #2
BvU
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Hollow conducting cylinder ? Height/diameter >> 1 ? Not grounded, I presume ?

Can any such imaginary closed surface be constructed in this case?
Any closed surface is a Gaussian surface, so: yes!
In short, will any flux be coming out of the closed end?
" 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 ...
 
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  • #3
Dale
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How much should be the flux coming out from the closed end?
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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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 ...
Let me clear it up a bit. Sorry for the confusing terminology.

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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BvU
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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.
Could you elaborate ?
 
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Dale
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Could you elaborate ?
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’ law
 
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  • #7
BvU
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How does one find ##\vec E\;## for such a configuration ?
 
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ZapperZ
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Say you have a hollow cylinder, whose one side is open. Now, you pace a positive charge ##Q## at the centre of this open end (such that it is just inside the cylinder). How much should be the flux coming out from the closed end?

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.
 
  • #9
Dale
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How does one find ##\vec E\;## for such a configuration ?
Coulomb’s law

Edit: oops, I see now that the cylinder is not just a partial Gaussian surface, but an actual conductor.
 
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  • #10
BvU
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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.
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

upload_2018-6-24_18-14-11.png
 

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  • #11
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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 me

View attachment 227263
Thanks for the diagram, I meant that.
 
  • #12
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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.
As per the diagram given by @BvU, one side of the cylinder is open. That's where the problem lies.
 
  • #13
ZapperZ
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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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  • #14
BvU
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Would the ##\vec E \;## field be distorted by the conducting can in such a way that the flux through the bottom is the same as in the case the can isn't there ?
 
  • #15
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Would the ##\vec E \;## field be distorted by the conducting can in such a way that the flux through the bottom is the same as in the case the can isn't there ?
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.

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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