Justification for 0 net E field within a charged shell

1. Sep 21, 2016

Old_sm0key

SORRY TITLE SHOULD READ: "JUSTIFICATION FOR 0 NET E FIELD WITHIN A CHARGED SHELL"

<< Title fixed by Moderator (except for the all-caps of course) >>

1. The problem statement, all variables and given/known data

A conceptual query:
Considering an irregular shaped conducing shell with a net charge, using Gauss' Law, any Gaussian surface located within the shell will enclosed 0 charge. Hence the flux integral vanishes.
But clearly these charges are radiating field lines within the shell, so each elemental flux, $\underline{E}.d\underline{A}$, will be non zero...? Thus why is that the LHS Gauss' Law would be 0?

2. Relevant equations
$\displaystyle\int_S\underline{E}.d\underline{A} = \frac{Q}{\epsilon_0}$

3. The attempt at a solution
(see part 1)

Last edited by a moderator: Sep 21, 2016
2. Sep 21, 2016

The Buttered Cat

Hi,

How about this question- I can just take a point charge, draw a Gaussian surface somewhere in space that does not enclose the charge, and the total flux through that shell is also zero, but you wouldn't say that the field due to that point charge is zero. What is the difference?

3. Sep 22, 2016

rude man

Think NET flux integrated over your inside Gaussian surface. Yes, the flux is very irregular but your flux integral will still be zero. Not obvious, to be sure, but the theorem can be proven by strictly mathematical means (the "Divergence" theorem) plus Maxwell's ∇⋅D = ρ.

Good that you're asking questions like this!

4. Sep 22, 2016

kuruman

A simpler method is this:
Since the shell is conducting (I assume "conducing" is a typo), it is an equipotential which means that the potential difference between any point A and any other point B on the surface is zero. Now imagine a hypothetical field line that starts at A and ends at B going through the inside of the conductor. If that's the case, then the line integral $\int \vec {E} \cdot \vec{ds}$ cannot be zero if you follow a path along this hypothetical field line. Furthermore, the potential difference between A and B, which is the negative of the line integral, would also not be zero. We have a contradiction, which can only be resolved by accepting that there are no field lines inside the conductor.

And a high-powered method (like killing a gnat with a sledgehammer) if you are familiar with the uniqueness theorem:
Inside the conductor, V = constant is a solution to Laplace's equation and satisfies the boundary conditions. Therefore, it is the solution. Since the electric field is the gradient of the potential, the electric field is zero inside the conductor.

5. Sep 22, 2016

Old_sm0key

Thanks all for the input. Will have a proper ponder tomorrow when more awake!

Thought it useful to clarify explain my situation: I'm just starting the second year of a physics masters degree in Britain, and so a couple of things mentioned e.g. that Maxwell equation, I look forward to covering later in the year.

I've read various posts in the past, but decided to take the plunge and join the forum to ask my own questions, starting with this post. Glad that I did, and I value the effort that you've all put into helping aspiring physicists like myself!

6. Sep 22, 2016

kuruman

Glad to have you in our community.

7. Sep 22, 2016

SammyS

Staff Emeritus
The lack of Electric Field, is caused primarily because you're dealing with a conductor, rather than being due to some application of Gauss's Law.

Let's suppose that you can temporarily set up an electric field within some conducting material. There are charges which are free to move. (In a metal those are negative charges but the sign doesn't really matter.) Negative charge will move in a direction opposite the field. (Any positive charges would move in the same direction as the field.) The electrons move until they reach the boundary of the conductor leaving excess negative charge there, and they are drawn away from as far as the opposite boundary of the conductor, leaving a net positive charge there. This process continues until those excess charges at the boundary produce a field cancelling the originally imposed field. This all happens relatively quickly. Thus any internal field is cancelled, unless some external agent keeps removing the excess electrons from the negative boundary and keeps replenishing them at the positive boundary, in which case we have current flow. - The simplest model for conduction of current in a wire - but that's not electrostatics.

8. Sep 22, 2016

rude man

Best of fortune!
PS post #4 is well worth pondering!

9. Sep 22, 2016

rude man

The line integral $\int \vec {E} \cdot \vec{ds}$ is of course zero but does that mean that E is zero everywhere along that path? (I know it is but is it a valid argument?)
E.g. ampere's law gives the integral around any closed path as current piercing that path but that does not imply H is zero everywhere along that path unless there is appropriate symmetry.

10. Sep 22, 2016

kuruman

Given: The integral is zero along the path that follows a hypothetical field line inside the conductor connecting points A and B on the surface. If the E-field were not zero everywhere, then it follows that the integral is positive along a portion of the path from A to B and negative along another portion in such a way that the positive and negative contributions to the line integral add to exactly zero. OK, but that implies that the E-field changes direction at least at one point along the path. This, in turn, implies a non-zero divergence of E at that point, i.e. some charge inside the conductor. That cannot be (in the static case) because Gauss's law has already told us that, if we place a charge inside a conductor, when equilibrium is reached, all the free charge is on the surface.

The analogy with the magnetic field is not appropriate because magnetic field lines cannot reverse direction at a point. The divergence of B is zero but the divergence of E is not.

11. Sep 22, 2016

rude man

That sounds right. Thanks.