- #1

- 2,138

- 2,713

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?

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter Wrichik Basu
- Start date

In summary: 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...

- #1

- 2,138

- 2,713

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?

Physics news on Phys.org

- #2

BvU

Science Advisor

Homework Helper

- 16,075

- 4,852

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 ...

- #3

Dale

Mentor

- 35,696

- 14,077

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?

- #4

- 2,138

- 2,713

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.

- #5

BvU

Science Advisor

Homework Helper

- 16,075

- 4,852

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.

- #6

Dale

Mentor

- 35,696

- 14,077

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 ?

- #7

BvU

Science Advisor

Homework Helper

- 16,075

- 4,852

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

- #8

- 32,820

- 4,719

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.

- #9

Dale

Mentor

- 35,696

- 14,077

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.

Last edited:

- #10

BvU

Science Advisor

Homework Helper

- 16,075

- 4,852

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.

- #11

- 2,138

- 2,713

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

- #12

- 2,138

- 2,713

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.

- #13

- 32,820

- 4,719

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.

- #14

BvU

Science Advisor

Homework Helper

- 16,075

- 4,852

- #15

Dale

Mentor

- 35,696

- 14,077

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.

Gauss' Law is a fundamental law of electromagnetism that relates the electric flux through a closed surface to the charge enclosed by that surface. It is named after the German mathematician and physicist, Carl Friedrich Gauss.

To find the electric flux using Gauss' Law, you first need to determine the electric field at every point on the closed surface. Then, you can calculate the total electric flux by integrating the dot product of the electric field and the surface area vector over the entire surface. The formula for electric flux using Gauss' Law is Φ = ∫E⃗ · dA⃗.

Electric flux is directly proportional to the electric field. This means that as the electric field increases, the electric flux also increases. They are also related by the angle between the electric field and the surface, with the maximum electric flux occurring when the electric field is perpendicular to the surface.

Yes, Gauss' Law can be used to find electric flux for any shape of closed surface as long as the electric field and the surface are both symmetric. This means that the electric field must have the same magnitude and direction at every point on the surface, and the surface itself must be symmetrical in shape.

The units of electric flux are N·m²/C or V·m, which are equivalent to coulombs per square meter (C/m²). This can also be written as volts per meter (V/m), as electric flux is also a measure of the electric field passing through a given area.

- Replies
- 83

- Views
- 3K

- Replies
- 1

- Views
- 351

- Replies
- 30

- Views
- 2K

- Replies
- 5

- Views
- 3K

- Replies
- 2

- Views
- 371

- Replies
- 8

- Views
- 2K

- Replies
- 1

- Views
- 1K

- Replies
- 5

- Views
- 2K

- Replies
- 25

- Views
- 2K

- Replies
- 5

- Views
- 2K

Share: