Rings of Electric Field and Gauss's Law

[flyingpig]

Homework Statement

Trying to find the E-field inside a conductor using rings even though my book tells me it is 0.

I haven't learned how to do surface integrals yet but I think I only need Calc II to do this.

The Attempt at a Solution

[PLAIN]http://img18.imageshack.us/img18/699/unledoj.jpg

So one ring produces a E-field of

[tex]E = \frac{kq}{x^2 + a^2}[/tex]

Now graphing [tex]x^2 + y^2 = a^2[/tex]

[PLAIN]http://img713.imageshack.us/img713/9007/unledbz.jpg

Now what I want to do is to take cross sections of my rings with the E-field up there and sum it up like a volume.

Now the problem is, I am not sure how to set it up properly

I tried doing

[tex]dE = \frac{-2xkq}{x^2 + a^2} dx[/tex]

[tex]dx = \frac{x^2 + a^2}{-2xkq} dE[/tex]

Then I tried doing

[tex]A(x) = \pi r^2 = \pi (x^2 + a^2)[/tex]

And hence

[tex]V = \int \frac{\pi (x^2 + a^2)^2}{-2xkq} dE[/tex]

Now the problem is, what aer the bounds? Did I even set up my integral right?

I watched a video and it says that the E-field will point in opposite directions but same magnitude, but I don't understand why is that so.

I know that if my bounds are +E and -E, I would get 0 for the integral because my function is odd. Now the problem is, how do I distinguish this from an insulator? I could've done the exact same thing for an insulator and they would both come out 0

 

[Idoubt]

your integration would be much simpler if you split the electric field into two components, one pointing towards/away the center of the right and one parallel to the plane of the ring.

 

[flyingpig]

your integration would be much simpler if you split the electric field into two components, one pointing towards/away the center of the right and one parallel to the plane of the ring.

What dose that mean...?

 

[Idoubt]

http://img685.imageshack.us/img685/7199/electricfield.jpg

Uploaded with ImageShack.usIn the image you can see that only the cos component has an effect, because for every element dl there will be a diametrically opposite element dl which has dEsin(theta) in the opposite direction, so dEsin(theta) will vanish in the integration and you now only need to worry about the cos component.

Now you can calculate dEcos(theta) for the element dl and integrate around the circle and you will get your electric field. ( hint : charge of the element is linear charge density times dl)

 

[Idoubt]

your integration would be much simpler if you split the electric field into two components, one pointing towards/away the center of the right and one parallel to the plane of the ring.

sorry that's :

one pointing towards/away the center of the RING and one parallel to the plane of the ringfor the rest of your question, what shape is your conductor?

 

[flyingpig]

Probably a spherical conductor, what happens if I want to do the same thing for an insulator?

 

[flyingpig]

http://img685.imageshack.us/img685/7199/electricfield.jpg

Uploaded with ImageShack.us


In the image you can see that only the cos component has an effect, because for every element dl there will be a diametrically opposite element dl which has dEsin(theta) in the opposite direction, so dEsin(theta) will vanish in the integration and you now only need to worry about the cos component.

Yeah I knew that, but I wanted to do the magnitude for a "general" case, but that could too.

Now you can calculate dEcos(theta) for the element dl and integrate around the circle and you will get your electric field. ( hint : charge of the element is linear charge density times dl)

Why dl? If we are only concerned with the horizontal direction, why not dx?

Also, what are my limits of integration? How do I know to have -E to E for conductor? WHY does it point opposite from the other symmetric end?

 

[Idoubt]

Yeah I knew that, but I wanted to do the magnitude for a "general" case, but that could too.

If you integrate, you will get field as E = KQx/ (a²+x²)^3/2 for a charged ring.

Why dl? If we are only concerned with the horizontal direction, why not dx?

Because the element dl is causing that field dE, so you need to add up all the dEs caused by each dl, which means you find the electric field dE for one dl and integrate from L=0 to L=max

( dl can be assumed to be a point charge and you can use coloumb's law)

Also, what are my limits of integration? How do I know to have -E to E for conductor? WHY does it point opposite from the other symmetric end?

the limits are the limits of the length of the conductor right? if we say L is the length of the conductor, where does it start and end?

http://img135.imageshack.us/img135/7199/electricfield.jpg

Uploaded with ImageShack.us

for a diametrically opposite element dl^, the sin component is oppsite

How do I know to have -E to E for conductor? - I didn't understand this.

 

[flyingpig]

-E because it points the other direction...?

 

[Idoubt]

-E because it points the other direction...?

I'm sorry I'm still confused, do you mean the small electric field dE due to an element dl or the total electric field E?

 

[flyingpig]

The total E

 

[Idoubt]

If you accept that the dEsin() components will cancel due to the symmetry of the problem, then the total electric field is the sum of all the dEcos() due to each dl, and dEcos() always points in the x direction so you can just integrate without worrying about the directions ( they are vectors after all )

Here is a derivation of this see if you get it

http://img171.imageshack.us/img171/682/giaf.jpg

Uploaded with ImageShack.us

 

[Trifis]

I'd like to know WHY it's impossible to solve this problem with Gauß's law!

 

[Idoubt]

To use gauss's law you need some sort of symmetry. That is you need a surface on which the electric field is a constant.

If E is constant then you can take it out of the integral [itex]\oint E.ds[/itex]

The problem here is that there is no such surface. The only places the E field would be constant will be concentric circles, but we need closed surfaces to use guass's law.

 

[Trifis]

@Idoubt thanks for the reply!

I was hoping though for a more profound answer, having regard to actually why there is no such symmetry in this and other similar charge distributions! Which geometrical or topological criterion decides after all the applicability of this specific law?!

 

[Idoubt]

well I don't think there is any profound answer. The power of symmetry arguments is very simple, if you rotate something about it's axis of symmetry then it is indistiguishible from it's previous state eg : a sphere by any angle, block by 90deg etc. The clinching argument is that the if you cannot see any change in the shape, orientation or distance of a charged body by moving on a particular surface, the electric field should be the same for the whole surface. This is also why we need infinite line charges and planes to use gauss's law..otherwise if you move a little from the centre , there is more charge on one side than the other

Guass's law is always correct but it is simply useless without some symmetry, I.E all guass law problems depend on taking the electric field outside the integral, if we can't do that its basically incalculable.

 

[Trifis]

Ok I have thought a lot about it and I found what truly is my problem here... I need a mathematical proof that should indicate the following: The direction of the elcetric field must be radial, for a charge distribution to remain invariant after applying a rotation matrix to its field.

In other words, why is the electric field of the symmetric objects the way it is? Is it an experimental deduction or just the result of a superposition?
I need maths in order to unravel this symmetry...

 

[Idoubt]

As for the maths, I think its a bit beyond me. If we were talking about a radial field like a point charge, the most obvious way to look for a symmerty is to look for a cyclic co-ordiante. The derivatives w.r.t to θ and [itex]\varphi[/itex] ( polar and azimuthal angles in sperical polar co-ords ) should be zero. If you think about it , what we mean by spherical symmerty is simply that it does not depend on an angular change, hence a pure radial dependence.

 

[Trifis]

@Idoubt thanks for the replies but "phyical languages" don't satisfy me anymore... The fact that the field of a point charge has a radial symmetry is an experimental assessment and doesn't indicate that the same should happen for an infinite number of point charges directed in space according to a special geometry...

I think I should pass this question on another thread ;)

 

[Idoubt]

good luck with your question. But i think it might be tough to find a fundamental mathematical reasoning. after all mathematical models are simply the most efficient form of physical arguments and generally we base them on experimental results. So I have always felt math comes after physics :)