Is My Understanding of Gauss's Law and Flux Contradictory?

[flyingpig]

Homework Statement

My book says

In two situations, there is zero flux through a closed surface: either

(1) There are no charged particles enclosed by the surface

(2) There are charged particles enclosed, but the net charge inside the surface is zero.

For either case it is incorrect to conclude that the electric field on the surface is zero. Gauss's law states that the electric flux is proportional to the enclosed charge, not the electric field.

Now here is the thing I have two situations

(1) Consider a hollow conducting sphere with a net charge +Q and a conducting solid sphere with a net charge -Q. The solid one is inside the hollow conducting sphere. If I were to construct a Gaussian Surface over the two spheres, I get a net charge of 0. Normally in my HW problems we would just say the E-field is 0, but according to the book (case 2) this is wrong. Why?

(2) Consider just a solid conducting sphere with a charge of +Q. The net charge inside must be 0 because we are inside a conductor. I think most books got it wrong when they say the E-field is zero because there are no charges inside, clearly there are free electrons that are free to roam, but their net charge (with protons) cancel them out to make a net E-field of 0. Perhaps the excess protons and electrons go on about on the surface and all them also cancel each other until the other charges reside on the surface (hence we have terms like a net charge of negative)

Now if I were to construct a Gaussian Surface inside my conductor, I get a net charge closed (most books leave out the summation sign and really confuses me, I think it's important to not exclude it) of 0. We could conclude it is 0, but according to Case 2, I cannot.

So what is going on here?

 

[Doc Al]

(1) Consider a hollow conducting sphere with a net charge +Q and a conducting solid sphere with a net charge -Q. The solid one is inside the hollow conducting sphere. If I were to construct a Gaussian Surface over the two spheres, I get a net charge of 0. Normally in my HW problems we would just say the E-field is 0, but according to the book (case 2) this is wrong. Why?

Just because case 2 is met (the net charge is zero) does not immediately allow you to deduce that the field is zero. Only if additional symmetry exists can you make that claim.

(2) Consider just a solid conducting sphere with a charge of +Q. The net charge inside must be 0 because we are inside a conductor. I think most books got it wrong when they say the E-field is zero because there are no charges inside, clearly there are free electrons that are free to roam, but their net charge (with protons) cancel them out to make a net E-field of 0. Perhaps the excess protons and electrons go on about on the surface and all them also cancel each other until the other charges reside on the surface (hence we have terms like a net charge of negative)

Now if I were to construct a Gaussian Surface inside my conductor, I get a net charge closed (most books leave out the summation sign and really confuses me, I think it's important to not exclude it) of 0. We could conclude it is 0, but according to Case 2, I cannot.

Again, just because you meet the minimal condition of case 2 does not allow you to immediately conclude that the field is zero. You need more to conclude that the field is everywhere zero in a conductor.

Don't confuse: Condition A does allow you to conclude B
With: Condition A contradicts B

In other words: Gauss's law tells you about the flux. You need additional info (symmetry, for instance) to use that information to find the field.

 

[flyingpig]

But it is 0. Let's just look at my second situation, inside a conductor for instance

\frac{\sum_{i=1}^{n} Q_1 + (-Q_2) +Q_3+...\pm Q_n}{\epsilon_0} = \oint{\vec{E}\cdot \vec{dA}}

0 = \vec{E} (4\pi r^2)

Anyways in all of my problems all surfaces have symmetry.

Where r cannot be zero because we know that our Gaussian Surface isn't a dot, so E must be 0

 

[flyingpig]

Doc Al, I dug this old thread https://www.physicsforums.com/showthread.php?t=225552&page=2

In electrostatic equilibrium, the field within a conductor is zero. Regardless of shape. Don't think so? Tell us why.

Which contradicts (at least to me)

In other words: Gauss's law tells you about the flux. You need additional info (symmetry, for instance)to use that information to find the field.

If you think you have shown that electrostatic equilibrium does not imply zero electric field everywhere within a conductor, you would be wrong.

I know zero flux itself does not mean the absence of a E-field, but the way the book says seems like it's saying that the E-field will never be zero.

 

[SammyS]

Homework Statement

My book says

"In two situations, there is zero flux through a closed surface: either

(1) There are no charged particles enclosed by the surface

(2) There are charged particles enclosed, but the net charge inside the surface is zero.

For either case it is incorrect to conclude that the electric field on the surface is zero. Gauss's law states that the electric flux is proportional to the enclosed charge, not the electric field.​

Now here is the thing I have two situations

(1) Consider a hollow conducting sphere with a net charge +Q and a conducting solid sphere with a net charge -Q. The solid one is inside the hollow conducting sphere. If I were to construct a Gaussian Surface over the two spheres, (I assume you mean the Gaussian surface is external to the spherical shell.) I get a net charge of 0. Normally in my HW problems we would just say the E-field is 0, but according to the book (case 2) this is wrong. Why?
...

So what is going on here?

Your text is discussing some cautions to keep in mind when using Gauss's Law.

You canNOT conclude that the E field is zero on your Gaussian surface (in your example 1) using only Gauss's Law - at least not by invoking Gauss's Law only on the single Gaussian surface you constructed.

In fact, if you place an additional charge of q ( not necessarily equal to Q) outside of your Gaussian surface, the electric flux through the Gaussian surface would still be zero, but E would be non-zero over all of your Gaussian surface.

 

[Doc Al]

Doc Al, I dug this old thread https://www.physicsforums.com/showthread.php?t=225552&page=2

In electrostatic equilibrium, the field within a conductor is zero. Regardless of shape. Don't think so? Tell us why.

Which contradicts (at least to me)

In other words: Gauss's law tells you about the flux. You need additional info (symmetry, for instance)to use that information to find the field.

Where's the contradiction? You seem to think that "Gauss's law alone does not imply zero electric field" means that "the electric field can never be zero". Why in the world would you think that?

I know zero flux itself does not mean the absence of a E-field,

It's not clear that you do. Realize that Gauss's law talks about the flux.

but the way the book says seems like it's saying that the E-field will never be zero.

The quote from your book in your first post does not say that. For the two cases they mention, you should be able to think of situations where the field is not zero.

You seem to be stuck thinking that "it is incorrect to conclude X" means that "X cannot be true". This is more a question of semantics and logic, than physics.

 

[flyingpig]

Where's the contradiction? You seem to think that "Gauss's law alone does not imply zero electric field" means that "the electric field can never be zero". Why in the world would you think that?

My logic is flawed...

Reading it again, Gauss's Law does indeed tell us inside the conductor the field is 0, there is doubt. What the book is trying to warn me is that do not think that the there isn't a field outside my surface, is that what they trying to tell me by on the surface?