# Application of Gauss' law -- some doubts

#### logearav

Joined Aug 19, 2011
243
Revered Members,
I have attached images of applications of Gauss' law namely 1) Electric field due to an infinitely long charged wire and 2) Electric field due to an infinite charged plane sheet.
In both cases Gaussian surface is cylinder.
In the first case, the electric flux is found for curved surface and it has been done for only side that is right side of cylinder in this case
In the second case, the electric flux is found for plane caps but it has been done for both the left plane cap and right plane cap.
My doubt is , why we are doing for only one side for the curved surface and for both sides in the case of plane caps?

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#### joeyd999

Joined Jun 6, 2011
4,477
Revered Members,
I have attached images of applications of Gauss' law namely 1) Electric field due to an infinitely long charged wire and 2) Electric field due to an infinite charged plane sheet.
In both cases Gaussian surface is cylinder.
In the first case, the electric flux is found for curved surface and it has been done for only side that is right side of cylinder in this case
In the second case, the electric flux is found for plane caps but it has been done for both the left plane cap and right plane cap.
My doubt is , why we are doing for only one side for the curved surface and for both sides in the case of plane caps?
My understanding is, and this goes back 25 years, that due to symmetry, you may choose any geometry you like that helps to simplify the problem. As for these particular cases, I'm not going to try to analyze them, but I assume the author chose his methods for the easiest approach to the solution.

#### davebee

Joined Oct 22, 2008
540
For the first case, the drawing shows a small surface segment to the right side of the cylinder, but the integration operation sums the contribution of all such surface segments all around the entire curved surface, then multiplies by the height to get the full area.

The drawing is just to show the meaning of all the variables.

#### logearav

Joined Aug 19, 2011
243
Thanks davebee. I understood your explanation. But why the integration is done twice in the second case, one for the right plane cap and again for the left plane cap? Does a single integration can't give the answer? We are doing only one integration for the entire curved surface in the first case.

#### joeyd999

Joined Jun 6, 2011
4,477
Thanks davebee. I understood your explanation. But why the integration is done twice in the second case, one for the right plane cap and again for the left plane cap? Does a single integration can't give the answer? We are doing only one integration for the entire curved surface in the first case.
Because you need to enclose the charge for the equations to work! In the first case, the cylinder encloses the charge for the length of wire. In the second, The two cylinders enclose the charge on the sheet. Instead of thinking 2 cylinders, think of it as one cylinder bisected by the sheet.

Keep in mind that the field lines in the first case are perpendicular to the cylinder wall. All field lines intersect the wall. In the second, they are perpendicular to the end-cap planes. 1/2 of the field lines intersect each of the end caps.

#### logearav

Joined Aug 19, 2011
243
Sorry joeyd999, I can't understand.

#### joeyd999

Joined Jun 6, 2011
4,477
Sorry joeyd999, I can't understand.
To simplify: in the first case, the field is being projected onto a single surface (the cylinder). In the second, the field is being projected onto two surfaces (the end caps of the bisected cylinder). If you only do a single integration, you are only taking into account 1/2 of the field.

To complicate: Let's say in the first case, you surrounded the wire with a cube, instead of a cylinder. 4 of the faces would intersect the field, so you would need 4 integrations instead of one. (Don't do this...the math would be atrocious, and the result would be useless -- unless you wanted to know the flux density at a point on one of the surfaces).

Remember, these equations begin with the premise that (IIRC):

surface integral of E dot dA =4 * pi * k * Qenclosed.

The charge must be enclosed for a valid result. Sorry, I don't have time to learn TeX for a better looking equation.

#### joeyd999

Joined Jun 6, 2011
4,477
BTW, back in college (many years ago), I couldn't comprehend Maxwell's equations until after a night of heavy drinking and a long, hot shower before my exam the next morning!

#### davebee

Joined Oct 22, 2008
540
This does not need to be so difficult. All that is being done is integrating an area based on the flux strength at that area. The math may get a little complicated when you have to integrate over different surfaces like cylinders or circles or squares, but the overall concept should not be that difficult.

For the case of the two ends, they could have claimed symmetry and only integrated over one of them and multiplied the result by two. Same if the surface was a cube; again, by symmetry, the result would be four times the sum over one face, and all you'd have to do is integrate over that one face. And even that should not be so hard; you just express the flux over a small area based on the distance to that area and the angle that the flux makes to the area, then integrate that expression over the face of the cube.

Is the math obscuring the physics? The physics is just saying that the flux over the whole area is the sum of the flux over every small portion of the area, and the math is what does the adding of the small parts, based on the flux strength over that small area.

#### logearav

Joined Aug 19, 2011
243
@joeyd999, Thanks a lot for your painstaking effort. In the first case also, there is E projecting towards left side and also to the right side, similar to second case. Thats what my doubt is?
@davebee, Thanks a lot for your beautiful explanation. Why can't we multiply by flux of Curved Surface Area of cylinder(first case) by two , which gives the result
E*4*pi*r*l?

#### joeyd999

Joined Jun 6, 2011
4,477
@joeyd999, Thanks a lot for your painstaking effort. In the first case also, there is E projecting towards left side and also to the right side, similar to second case. Thats what my doubt is?
No. In the first case, the field lines are being projected radially in all directions (not just left and right), perpendicular to the entire surface of the cylinder. The single integration covers the entire surface.

#### davebee

Joined Oct 22, 2008
540
@davebee, Thanks a lot for your beautiful explanation. Why can't we multiply by flux of Curved Surface Area of cylinder(first case) by two , which gives the result
E*4*pi*r*l?
you could, but why would you do that? You're asking a math question that shows that you don't understand the physics of the problem.

Compare these two problems. Both use cylinders. In the first case, electric flux intersects with the curved cylinder wall but not the two ends. In the second case, electric flux intersects with the two ends but not the curved cylinder wall.

In the first case, you integrate the expression for flux strength over the curved surface. In the second, you integrate the expression over the two ends.

One way to solve the second case would be to perform two integrations then add the results, as in the original solution. Another way to solve it could be to observe that the surfaces have the same area and are at the same distance from the charges, so it would be equivalent to perform one integration then multiply the result by two.

That's where the "multiply by two" comes from, which makes no sense for the first case.