Conductor shielding in electrostatics

[daudaudaudau]

Homework Statement

Use Gauss' theorem (and the fact that the line integral of the electric field around a closed loop is zero, if necessary) to prove that a closed, hollow conductor shields its interior from fields due to charges outside. (This is 1.1b in Jackson, and it's not really homework, but I guess this is still the right place).

The Attempt at a Solution

So we're trying to prove that whatever field is outside the conductor, the field inside the hollow part is always zero. The solution I have seen various places draw a contour which is half inside the hollow part and half inside the conductor, and the line integral is then \int_{\mathcal C}E_{\text{hollow}}\cdot d\ell=0, because the E-field is zero in the conductor. At this point, how can we conclude that E_{\text{hollow}} is zero? Might it not be positive some places and negative in other places, giving a zero integral?

 

[kuruman]

Homework Statement

Use Gauss' theorem (and the fact that the line integral of the electric field around a closed loop is zero, if necessary) to prove that a closed, hollow conductor shields its interior from fields due to charges outside. (This is 1.1b in Jackson, and it's not really homework, but I guess this is still the right place).

The Attempt at a Solution

So we're trying to prove that whatever field is outside the conductor, the field inside the hollow part is always zero. The solution I have seen various places draw a contour which is half inside the hollow part and half inside the conductor, and the line integral is then \int_{\mathcal C}E_{\text{hollow}}\cdot d\ell=0, because the E-field is zero in the conductor. At this point, how can we conclude that E_{\text{hollow}} is zero? Might it not be positive some places and negative in other places, giving a zero integral?

There is a certain subtlety in the argument. The line integral for a path entirely inside the conductor is zero because the electric field is zero. Now consider a path that starts and ends at the same points, but is entirely in the hollow and follows along one of the assumed E field lines. In this case E^.dl is always positive and when you integrate, you add a whole lot of positive numbers. Thus the line integral seems to depend on the path. For this to be true, the electric field must be non-conservative. Since the curl of E is zero, the field is conservative so it must be zero.

 

[daudaudaudau]

There is a certain subtlety in the argument. The line integral for a path entirely inside the conductor is zero because the electric field is zero. Now consider a path that starts and ends at the same points, but is entirely in the hollow and follows along one of the assumed E field lines. In this case E^.dl is always positive and when you integrate, you add a whole lot of positive numbers. Thus the line integral seems to depend on the path. For this to be true, the electric field must be non-conservative. Since the curl of E is zero, the field is conservative so it must be zero.

Can we not do this without referring to concepts such as field lines? I.e. using only Gauss' theorem and the conservative nature of the E-field? I guess all we know is that the surface integral of the charge density on the inside of the conductor is zero \oint_\mathcal{S} \rho_s ds=0, and then we need to prove that the E-field is zero, i.e. E(\mathbf r)=\oint_\mathcal{S'}\rho_s(\mathbf r')\frac{\mathbf r-\mathbf r'}{|\mathbf r-\mathbf r'|^3}ds=0.

 

[gabbagabbahey]

The solution I have seen various places draw a contour which is half inside the hollow part and half inside the conductor, and the line integral is then \int_{\mathcal C}E_{\text{hollow}}\cdot d\ell=0, because the E-field is zero in the conductor. At this point, how can we conclude that E_{\text{hollow}} is zero? Might it not be positive some places and negative in other places, giving a zero integral?

The point is that you can choose any path \mathcal{C} inside the hollow cavity and it will be true that \int_{\mathcal C}E_{\text{hollow}}\cdot d\ell=0...You can even make the path infinitesimally small so that E_{\text{hollow}} has a single value along the path...the only way for the integral to always be zero, regardless of the path is that the field itself is zero.

 

[daudaudaudau]

The point is that you can choose any path \mathcal{C} inside the hollow cavity and it will be true that \int_{\mathcal C}E_{\text{hollow}}\cdot d\ell=0...You can even make the path infinitesimally small so that E_{\text{hollow}} has a single value along the path...the only way for the integral to always be zero, regardless of the path is that the field itself is zero.

You cannot choose any path. The path must start at the conductor and end at the conductor. Think about a uniformly charged sphere with no holes. All line integrals starting and ending on the sphere are also zero, but this doesn't imply that there is no field outside the sphere.

 

[kuruman]

Can we not do this without referring to concepts such as field lines? I.e. using only Gauss' theorem and the conservative nature of the E-field? I guess all we know is that the surface integral of the charge density on the inside of the conductor is zero \oint_\mathcal{S} \rho_s ds=0, and then we need to prove that the E-field is zero, i.e. E(\mathbf r)=\oint_\mathcal{S'}\rho_s(\mathbf r')\frac{\mathbf r-\mathbf r'}{|\mathbf r-\mathbf r'|^3}ds=0.

If the concept of field lines bothers you, how about this argument:

I know that the conductor is at constant potential, say V₀. I assert that the potential inside the cavity is V₀ everywhere. This means that there is no E field inside the cavity.

Is my assertion justified? Well, V₀ (a constant) is a solution of Laplace's equation (valid inside the cavity because there are no charges there) and reduces to the appropriate value V₀ at the boundary (surface of conductor). Therefore, by the power vested in me by the Uniqueness Theorem, V₀ is indeed the solution inside the cavity. End of discussion.

Is that better?

 

[daudaudaudau]

Yeah I guess that works because the potential is continuous across the boundary, but I think it goes a bit beyond what the problem stated we could use to solve it. If you could only make the "field lines"-argument a bit more mathematical and less intuitive...

 

[kuruman]

I gave you the gist of the argument in my first post. Try to say what I said in English using the language of mathematics. I used field lines only to make sure that the line integral has positive contributions from beginning to end.