Gauss' theorem due to thin infinitely long sheet

[Rishabh076]

Our teacher derived an expression for electric field due to thin sheet using gauss law which was sigma/2*epsilon naught which was independent of r but dies that mean the field will be same even if r is 10000km but field is also equal to kq/r^2

 

[Rishabh076]

[PLAIN]http://pg [/PLAIN] 3.36 sl arora example 66[/PLAIN] [PLAIN]http://pg [/PLAIN] 3.36 sl arora example 66[/PLAIN] http://postimg.org/image/oeab0o9zt/http://postimg.org/image/oeab0o9zt/

 

[Rishabh076]

http://postimg.org/image/n31cy3x57/

 

[Rishabh076]

I want to know why field is not dependent on distance when it usually is

 

[Rishabh076]

Does this mean field will be same even at huge distances please answer

 

[Qwertywerty]

Edit : To your question , no , electric field wouldn't be σ/2ε₀ for far away points .

 

[Rishabh076]

thonding to your own posts is not a good idea .

To your question , no , electric field wouldn't be σ/2ε₀ for far away points .

Well it was my first time on a forum I didn't know then but according to equation e does not depend on r but it should I also put link of pictures of derivation I can t seem to get how it will and won't depend on r at the same time

 

[Qwertywerty]

Well it was my first time on a forum I didn't know then but according to equation e does not depend on r but it should I also put link of pictures of derivation I can t seem to get how it will and won't depend on r at the same time

The electric field due to a disc , or any other plane sheet is actually dependent on the distance of a point from the sheet .

For example , electric field due to a charge disk is σx/2ε₀(1/x - 1/(x²+R²)^0.5 - where R is radius of the disc.

But you'll see , if you consider R>>x , then electric field at a distance x is approximately σ/2ε₀ .

Thus , it is only an approximation , nothing more .

 

[Rishabh076]

The electric field due to a disc , or any other plane sheet is actually dependent on the distance of a point from the sheet .

For example , electric field due to a charge disk is σx/2ε₀(1/x - 1/(x²+R²)^0.5 - where R is radius of the disc.

But you'll see , if you consider R>>x , then electric field at a distance x is approximately σ/2ε₀ .

Thus , it is only an approximation , nothing more .

The electric field due to a disc , or any other plane sheet is actually dependent on the distance of a point from the sheet .

For example , electric field due to a charge disk is σx/2ε₀(1/x - 1/(x²+R²)^0.5 - where R is radius of the disc.

But you'll see , if you consider R>>x , then electric field at a distance x is approximately σ/2ε₀ .

Thus , it is only an approximation , nothing more .

Thnx qwerty thnx so much can u cite the place where I can see the derivation for new formula so I can derive it once for practice and reference
Thnx qwerty thnx so much

 

[Qwertywerty]

Try this - https://www.google.co.in/url?sa=t&s...T8ICAc&usg=AFQjCNFCRuM7tkJAyC3le4WFxhLl7fpZsg

 

[Rishabh076]

Try this - https://www.google.co.in/url?sa=t&s...T8ICAc&usg=AFQjCNFCRuM7tkJAyC3le4WFxhLl7fpZsg

But isn't field e=kq/r(squared) but PDF sayed it is kq/r(cubed) and why isn't Gauss's law applicable for this problem

 

[Qwertywerty]

But isn't field e=kq/r(squared) but PDF sayed it is kq/r(cubed)

Which step ?

 

[Rishabh076]

Which step ?

3 rd step

 

[Qwertywerty]

If you are referring to step 4 dE , it is the electric field due to ring of radius R - formula is kQx/(x²+R²)^1.5 .
https://www.google.co.in/url?sa=t&s...da2AtY&usg=AFQjCNEEeswTJkjqY3PMmuJz82fJ4k410g

 

[Rishabh076]

Which step ?

It is r(cubed) in step 3 of previous article and in this article I didn't get why Gauss's law was 4πq whereas it is q/e in our school here k was not in equation for deriving gays law but my initial problem was same as the pillbox and infinite sheet problem at end because it gives adifferent result

 

[Qwertywerty]

But isn't field e=kq/r(squared) but PDF sayed it is kq/r(cubed) and why isn't Gauss's law applicable for this problem

I think it's starting to get confusing . I'll just answer your doubt - you can't use gauss law because you do not have symmetry here ( Which kind ? Refer to original derivation for x << R ) .

You said you want to try to derive it - so do it . Take a disc , take an element ( ring ) of radius r ( < R ) , and find electric field due to this at a distance x . Now integrate , and match with the value I have posted earlier .

 

[eltodesukane]

Our teacher derived an expression for electric field due to thin sheet using gauss law which was sigma/2*epsilon naught which was independent of r but dies that mean the field will be same even if r is 10000km but field is also equal to kq/r^2

--Yes.
In a way it is easy to see.
Consider a sideway view of the situation.
Move the test point N times further away, and zoom out the sideway view N times.
The situation looks exactly the same.

Consider a fixed solid angle from the test point to the sheet.
Move N times further away, the field from each sheet element is multiplied by 1/N^2, but the sheet area within the solid angle is multiplied by N^2.
So the field is independent of distance.

 

[orzyszpon]

The kq/r^2 is the field of a POINT charge. Here, we have an infinite sheet of charge composed of infinite number of point charges. The further you are above the plane (z) the more of the charges come into play (actually, the number grows like r^2, which is why the field stays constant = (kq/r^2)*r^2). Consider the analogy of a balloon (observation point) rising above the Earth (flat surface of point charges). Although the influence of anyone charge wanes as 1/r^2, the literal field of view (what you the observer in the balloon sees) grows like r^2 as your "horizon" widens. This is not to advocate the flat Earth interpretation.
(actually, if your balloon could fly into space, the Earth would eventually become like a point charge).

 

[Qwertywerty]

--Yes.
In a way it is easy to see.
Consider a sideway view of the situation.
Move the test point N times further away, and zoom out the sideway view N times.
The situation looks exactly the same.

Consider a fixed solid angle from the test point to the sheet.
Move N times further away, the field from each sheet element is multiplied by 1/N^2, but the sheet area within the solid angle is multiplied by N^2.
So the field is independent of distance.

This is wrong .

See post #8 .

 

[zelong]

This is wrong .

See post #8 .

it is not wrong. For infinitely large charged sheet, it's electric field is independent of distance. For understanding this case, you can imagine electric field for a capacitor first, "very close" to the sheet, electric field is almost uniform. Now, the sheet is infinitely large which means anywhere is "very close". Guess's law cannot be wrong since it is basically a geometrical theorem.

 

[Qwertywerty]

it is not wrong. For infinitely large charged sheet, it's electric field is independent of distance. For understanding this case, you can imagine electric field for a capacitor first, "very close" to the sheet, electric field is almost uniform. Now, the sheet is infinitely large which means anywhere is "very close". Guess's law cannot be wrong since it is basically a geometrical theorem.

Actually , it is . I never said Gauss's law is wrong - I have stated that electric field is not independent of distance .

Gauss's law is usually useful in finding the electric field in certain symmetrical conditions . Do you know the derivation for electric field due to a large plane sheet using Gauss's law ?

Also , you should understand that infinite is relative . For an area of 1 m² , a distance of 10^-10 m would allow you to consider the sheet as infinite . A distance of 1 m would not .

 

[rcgldr]

For an infinite sheet, there are two main calculus methods. One is to treat an infinite disc as an infinite number of rings, the other is to treat an infinite plane as an infinite number of infinitely long and thin rectangles. See post #2 in this thread:

https://www.physicsforums.com/threads/distance-from-plane.471647