# Finding electric flux using Gauss' Law

• B
Gold Member

## Main Question or Discussion Point

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?

Related Other Physics Topics News on Phys.org
BvU
Homework Helper
2019 Award
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 ...

• Simon Bridge
Dale
Mentor
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.

Gold Member
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.

• Dale
BvU
Homework Helper
2019 Award
You know the electric field from a point charge, so just integrate that over the closed end to get the flux.
Could you elaborate ?

Dale
Mentor
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

• Wrichik Basu
BvU
Homework Helper
2019 Award
How does one find $\vec E\;$ for such a configuration ?

ZapperZ
Staff Emeritus
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.

Dale
Mentor
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.

Last edited:
BvU
Homework Helper
2019 Award
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 #### Attachments

• 1.3 KB Views: 511
• Wrichik Basu
Gold Member
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.

Gold Member
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.

ZapperZ
Staff Emeritus
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.

• Wrichik Basu
BvU
Homework Helper
2019 Award
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 ?

Dale
Mentor
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.

• Wrichik Basu and BvU