How is the uniqueness theorem is used in this example?

Question

I was going through example 2.10 in Griffiths (Introduction to Electrodynamics 4th edition), everything seemed okay, but at last he makes an assertion, the proof for which I am not able to understand.

Nope: As we'll see in the uniqueness theorems of Chapter 3, electrostatics is very stingy with its options; there is always precisely one way -- no more -- of distributing the charge on a conductor so as to make the field inside zero. Having found a possible way, we are guaranteed that no alternative exists, even in principle.

I searched for uniqueness theorems in Chapter 3, and I got two of them, but none of them seem to directly imply that there is always precisely one way- no more--of distributing the charge on a conductor so as to make the field inside zero.

First uniqueness theorem states that:

The solution to Laplace's equation in some volume $V$ is uniquely determined if $V$ is specified on the boundary surface $S$.

and the second one states that:

In a volume $V$ surrounded by conductors and containing a specified charge density $p$, the electric field is uniquely determined if the total charge on each conductor is given (Fig. 3.6). (The region as a whole can be bounded by another conductor, or else unbounded.)

My question: How is the assertion made in example 2.10 justified by the uniqueness theorems?

Answer 1

A paraphrase of this answer.

• The second uniqueness theorem states:

In a volume $V$ surrounded by conductors and containing a specified charge density $\rho$, the electric field is uniquely determined if the total charge on each conductor is given. (The region as a whole can be bounded by another conductor, or else unbounded.)

• Apply this theorem for the cavity region.
  The cavity as a whole is bounded by the conductor in question and there are no other conductors inside the cavity. The charge density inside the cavity is specified: $\rho(\mathbf{r})=q\delta^3(\mathbf{r}-\mathbf{r}_0)$ where $\mathbf{r}_0$ is the point where the charge $q$ is located.
  The theorem states that there is a unique electric field $\mathbf{E}(\mathbf{r})$ inside the cavity.
• This electric field $\mathbf{E}(\mathbf{r})$ also uniquely determines the surface charge distribution on the boundary surface (wall of the cavity) using the following electrostatic boundary condition (ref. eq. (2.33) in Griffiths):
  $$\mathbf{E}_{\text{above}}-\mathbf{E}_{\text{below}}=\frac{\sigma}{\epsilon_0}\hat{n} $$ "Above" and "Below" are with reference to which way the $\hat{n}$ points. Since the electric field inside the conductor is zero, $$\mathbf{E}(\mathbf{r'})=\frac{\sigma(\mathbf{r'})}{\epsilon_0} \hat{n}$$ where, $\mathbf{r'}$ is a point on the boundary surface and $\hat{n}$ is the inward pointing unit normal vector.
• One can see that the electric field inside the cavity and the surface charge distribution on the wall of the cavity are uniquely determined based on the shape of the cavity and the position of the charge inside the cavity alone. Everything else, such as the shape of the conductor that houses the cavity, is irrelevant.

Answer 2

It relates to the second uniqueness theorem.

In a volume $V$ surrounded by conductors and containing a specified charge density $\rho$, the electric field is uniquely determined if the total charge on each conductor is given (Fig. 3.6). (The region as a whole can be bounded by another conductor, or else unbounded.)

Therefore, if you specify the charge on the conductor, you are guaranteed that there is only one unique field configuration (and hence potential function (up to a constant)). By Gauss's law, this then means that there is only one unique way to distribute the charges on the conductor.

Then jump back to chapter 2. Griffiths has found one solution that works. Since we are guaranteed there is only one unique solution by the uniqueness theorem, this is the only solution.

This is the whole idea behind the third chapter. As long as you find some way to come up with a solution, you know that it is the solution. Even if it seems like a convoluted method such as the method of images, getting an infinite sum of separable solutions, etc. you know you got it, and you don't need to keep looking.

Answer 3

In a static situation, there can be no field inside a conductor. If there were, charges would move until there was no field. This means that every point within a conductor (including points on the surface surrounding a cavity) is at the same potential. With a fixed total charge, any rearrangement of the charge would change the potential at every point, and that would require a non-zero field. If there was a charge inside a cavity in the conductor, Gauss's law would require an equal and opposite charge on the inner surface of the conductor that surrounds the cavity.