Gauss' Law and gaussian surface

[EEWannabe]

Hey there, just had a question about Gauss' law, should be relativity simple however the explanation we were given was quite poor and only seems to apply well to the examples we were given. (This isn't homework).

I (think I ) know the equation for Gauss' Law and what it means, that basically, the net flux through a gaussian surface multiplied by the Electric Field strength at the gaussian surface is equal to the total enclosed charge by the surface all over epsilon_o.

However I'm having problems intuitively with that;

If you had a shape like this;

http://i27.photobucket.com/albums/c171/Chewbacc0r/gauss.jpg

Surely at that point where the Gaussian surface, there's obviously so charge enclosed but is there really no electric field? The only explanation I can think of is by thinking that charge wouldn't be uniformly distributed locally at that point, as it's much closer together, so the charge around the surface would be weaker?

That's possible, but I've got a feeling I've just misinterpreted the law somehow, if anyone could help that'd be great.

Edit; imagine the little structure at the bottom is at the middle of the box.

 

[berkeman]

Hey there, just had a question about Gauss' law, should be relativity simple however the explanation we were given was quite poor and only seems to apply well to the examples we were given. (This isn't homework).

I (think I ) know the equation for Gauss' Law and what it means, that basically, the net flux through a gaussian surface multiplied by the Electric Field strength at the gaussian surface is equal to the total enclosed charge by the surface all over epsilon_o.

However I'm having problems intuitively with that;

If you had a shape like this;

http://i27.photobucket.com/albums/c171/Chewbacc0r/gauss.jpg

Surely at that point where the Gaussian surface, there's obviously so charge enclosed but is there really no electric field? The only explanation I can think of is by thinking that charge wouldn't be uniformly distributed locally at that point, as it's much closer together, so the charge around the surface would be weaker?

That's possible, but I've got a feeling I've just misinterpreted the law somehow, if anyone could help that'd be great.

I'm not sure I understand your sketch. The key point about Gauss' law is that there will be no NET flux out of a closed surface unless there is some net charge enclosed by the surface. You can have lots of flux and an E-field, but the flux in one side of the surface equals the flux going out of the other side of the enclosed surface, unless net charge is enclosed.

 

[EEWannabe]

I'm not sure I understand your sketch. The key point about Gauss' law is that there will be no NET flux out of a closed surface unless there is some net charge enclosed by the surface. You can have lots of flux and an E-field, but the flux in one side of the surface equals the flux going out of the other side of the enclosed surface, unless net charge is enclosed.

Yeah I understand that, i'll re-draw the sketch as best I can, basically it's a 3d box where in the middle there's some sort of structure protruding into the middle of the box, at the red point, sureley if the charge is evenly distributed around the whole shape the E field at the red point would be non-zero, come to think of it, in almost any non-spherical shape this would be the case.

My question is that is Gauss' law true that there is infact no E-field inside? And would this follow for the inside of any hollow 3d shape? ( providing the charge is distributed on the outside and non inside) Or am I interpreting it wrongly? - Maybe is it something to do with how the charge is re-distributed to make the E field inside 0?

I'm really struggling to picture this and any tips on picturing it would be great.

 

[Doc Al]

My question is that is Gauss' law true that there is infact no E-field inside?

Gauss' law doesn't say that the field inside the volume will be zero. It says that the flux through the closed surface will be zero since there's no charge enclosed.

 

[EEWannabe]

Gauss' law doesn't say that the field inside the volume will be zero. It says that the flux through the closed surface will be zero since there's no charge enclosed.

Hmmm, I think I see what you're getting at,

So all gauss' law tells us is the relationship between the next flux and the enclosed charge? Does that then mean that really unless some clever symmetry is used Gauss' law can't be used to find the electric field stregnth inside any hollow 3d shape?

thanks for the replies

 

[Doc Al]

So all gauss' law tells us is the relationship between the next flux and the enclosed charge?

Right.

Does that then mean that really unless some clever symmetry is used Gauss' law can't be used to find the electric field stregnth inside any hollow 3d shape?

Exactly! While Gauss' law always applies, it's only useful for finding the electric field in special situations that have high symmetry. For an arbitrary shape, it's useless.

 

[EEWannabe]

Right.
While Gauss' law always applies, it's only useful for finding the electric field in special situations that have high symmetry. For an arbitrary shape, it's useless.

I think I understand now;

My teacher whenever they showed an example he jumped straight from the integral to the EA = Qenclosed / E_o, taking it for granted every time, since he said he "didn't want to get into the integration."

But I think I understand now that it the reason is related to the integral;

\oint E.dA

For the gaussian surface the Electric field is constant, for example in the sphere because the electric field is only dependent on distance from the centre, which is the same for the whole surface, and therefore can be taken out of the integral. To give EA = Qenclosed/Eo.

Whereas for a non-symmetrical example, such as the one above, by no means would it be constant, and you'd need the electrical field strength as a function of area or something else in order to evaluate it!

So is that right? Gauss' law can only be used where the electrical field strength is constant for the entire surface? - Or can at least be split up into numerous examples where that's the case?

Thanks again!

 

[Doc Al]

So is that right? Gauss' law can only be used where the electrical field strength is constant for the entire surface? - Or can at least be split up into numerous examples where that's the case?

You have it exactly right.

Here are the usual examples where Gauss' law can be used to find the field, since over the Gaussian surfaces used the electric field is constant (or zero): http://hyperphysics.phy-astr.gsu.edu/hbase/electric/gaulaw.html#c4"

 

[EEWannabe]

Thanks so much! You should be our teacher = D

 

[DaTario]

You have it exactly right.

Here are the usual examples where Gauss' law can be used to find the field, since over the Gaussian surfaces used the electric field is constant (or zero): http://hyperphysics.phy-astr.gsu.edu/hbase/electric/gaulaw.html#c4"

With all due respect I disagree. The main point on Gauss Law, related to this possible application, is that, in a certain number of cases, you are able to infer the field in one point from the knowledge of the flux of the electric field through some cleverly choosed surface. This does not imply that the field must be constant on the surface, although these situations are the most frequently found in exercises and the easiest to solve. The use of Gauss Law to provide the field value in one point may come, for example, from a situation where this geometrical point is a kind of point of mean electric field. So having the flux and the area, we may also deduce the value of E at the desired point.

Hoping to have reached some nice degree of clarity,

Best wishes

DaTario

 

[Doc Al]

With all due respect I disagree. The main point on Gauss Law, related to this possible application, is that, in a certain number of cases, you are able to infer the field in one point from the knowledge of the flux of the electric field through some cleverly choosed surface. This does not imply that the field must be constant on the surface, although this situations are the most frequently found in exercises and the easiest to solve.

Sure. As long as you can solve the integral to get the field, you're good. As you say, in most simple situations where Gauss' law is applied, the field is constant over the surface. Of course it doesn't have to be.

But the use of Gauss Law to provide the field value in one point may come, for example, from a situation where this geometrical point is a kind of point of mean electric field. So having the flux and the area, we may also deduce the value of E at the desired point.

Why don't you give an example?

 

[DaTario]

I will try, but it may take a week to prepare...

Best wishes

DaTario

 

[DaTario]

we may go first on some artificial setup, like this:

a point like charge Q is positioned in the origin of a catesian system of coordinates. Containing this charge there is a closed surface which passes through the point (1,2,5). We want to determine the field E0 at this point. We also know that the scalar product between the field and the normal unitary vector may be written as follows:

where E0, is the field value in (1,2,5).

Perhaps this example is good enough.

Best wishes

DaTario