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Following my instructor's notes the statement of the Uniqueness Theorem(s) are as follows

"If $\rho_{inside}$ and $\phi_{boundary}$ (OR $\frac{d \phi_{boundary}}{dn}$ ) are known then $\phi_{inside}$ is uniquely determined"

A few paragraphs later the notes state

"For the field inside S (a surface), knowing $\phi_{boundary}$ (OR $\frac{d \phi_{boundary}}{dn}$) everywhere on S is as good as knowing all the outside charges; it carries all the same information about their effects"

I don't see how this follows from the statement of the Uniqueness Theorem. If anything it seems to me that the instructor is saying the converse of the Uniqueness Theorem while flipping definitions of "inside" and "outside".

"If $\phi_{boundary}$ (OR $\frac{d \phi_{boundary}}{dn}$ ) are known on surface S then $\rho_{outside}$ is uniquely determined"

Can anyone help me

  1. decipher what my instructor is trying to say

  2. Offer help in the way of a formal proof or a convincing physical argument

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  • $\begingroup$ Is S in the second instance empty by any chance? $\endgroup$
    – noah
    Commented Mar 2, 2019 at 16:34
  • $\begingroup$ @noah The notes do not indicate. For the sake of argument let's suppose it is because I'm curious to hear what you have to say. $\endgroup$
    – Omar Azami
    Commented Mar 2, 2019 at 16:38

1 Answer 1

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"If $\phi_{\text{boundary}}$ (or $\frac{d \phi_{\text{boundary}}}{dn}$ ) is known on surface $S$, then $\rho_{\text{outside}}$ is uniquely determined"

This is not what the paragraph is trying to say.

"For the field inside $S$ (a surface), knowing $\phi_{\text{boundary}}$ (or $\frac{d \phi_{\text{boundary}}}{dn}$) everywhere on $S$ is as good as knowing all the outside charges; it carries all the same information about their effects"

It is saying that no matter what $\rho_\text{outside}$ exactly looks like, the only thing that matters for the field inside $S$ is the field it produces on the surface $S$. So it is pretty much exactly saying what your first paragraph tells you, minus the charges inside $S$. But from superposition it follows that the total field inside is determined by the influence of the outside charges (for which we only need to know $\phi$ on $S$), and the charges inside $S$, yielding the statement in your first paragraph.

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