B Symmetry regarding induced potentials?

[haruspex]

TL;DR Summary

Is it true that the potential arising at A from induced charges on conductor B from point charge q at C = that arising at C from induction on B from charge q at A?

A homework thread, https://www.physicsforums.com/threa...etal-sheet-along-a-spherical-surface.1057702/, references https://arxiv.org/pdf/1007.2175.pdf.
There is an uncharged conductor and a point charge. In the paper referenced, $\bar\phi_y(x)$ is defined as the potential at x due to the induced charges on the conductor when the point charge is at y.
As justification for eqn 10 it states that $\bar\phi_x(y)=\bar\phi_y(x)$.
I cannot see why that should be true, but I cannot construct any counterexample.
Is it evident, or maybe some standard result?

 

[pepos04]

Thank you for trying to clarify this point, I had already given up hope. But even if $\bar\phi_x(y)=\bar\phi_y(x)$ were true (assume this to be true for a moment, though it remains to be verified), what obvious relationship prompts the assertion that, since $bar\phi_x(y)=\bar\phi_y(x)$, then $\frac{q}{2} \nabla_x \phi_x(x) =- F(x)$?

 

[TSny]

As justification for eqn 10 it states that $\bar\phi_x(y)=\bar\phi_y(x)$.
I cannot see why that should be true, but I cannot construct any counterexample.
Is it evident, or maybe some standard result?

This is an interesting result that can be deduced from Green's Reciprocity Theorem (see equation 5).

 

[renormalize]

This is an interesting result that can be deduced from Green's Reciprocity Theorem (see equation 5).

Darn, you just beat me to it! But I'll go ahead and post my response:
Start from two charge densities $\rho_x,\rho_y$ that give rise to two electrostatic potentials $\phi_x,\phi_y$ via Poisson's equation:$$\nabla^{2}\phi_{x}\left(z\right)=-\rho_{x}\left(z\right)/\varepsilon,\;\nabla^{2}\phi_{y}\left(z\right)=-\rho_{y}\left(z\right)/\varepsilon$$and suppose that the charge densities and fields vanish sufficiently fast at spatial infinity so that boundary terms are ignorable (in other words, the usual physicist's assumption!). Then Green's second identity (https://en.wikipedia.org/wiki/Green's_identities) can be written in terms of $\phi_1,\phi_2$ as:$$0=\int\left(\phi_{x}\left(z\right)\nabla^{2}\phi_{y}\left(z\right)-\phi_{y}\left(z\right)\nabla^{2}\phi_{x}\left(z\right)\right)d^{3}z=\varepsilon^{-1}\int\left(\phi_{y}\left(z\right)\rho_{x}\left(z\right)-\phi_{x}\left(z\right)\rho_{y}\left(z\right)\right)d^{3}z$$where the integration extends over all space. This result is known in electrostatics as Green's Reciprocity (https://en.wikipedia.org/wiki/Reciprocity_(electromagnetism)). Now specialize to point charges $\rho_{x}\left(z\right)=q_{x}\delta^{3}\left(z-x\right),\;\rho_{y}\left(z\right)=q_{y}\delta^{3}\left(z-y\right)$ located at positions $x,y$:$$0=\varepsilon^{-1}\int\left(\phi_{y}\left(z\right)q_{x}\delta^{3}\left(z-x\right)-\phi_{x}\left(z\right)q_{y}\delta^{3}\left(z-y\right)\right)d^{3}z=\varepsilon^{-1}\left(q_{x}\phi_{y}\left(x\right)-q_{y}\phi_{x}\left(y\right)\right)$$In particular, if $q_x=q_y$ (as is apparently the case in the cited external reference), then the simple reciprocal relation $\phi_{y}\left(x\right)=\phi_{x}\left(y\right)$ indeed holds.

 

[TSny]

Now specialize to point charges $\rho_{x}\left(z\right)=q_{x}\delta^{3}\left(z-x\right),\;\rho_{y}\left(z\right)=q_{y}\delta^{3}\left(z-y\right)$ located at positions $x,y$:$$0=\varepsilon^{-1}\int\left(\phi_{y}\left(z\right)q_{x}\delta^{3}\left(z-x\right)-\phi_{x}\left(z\right)q_{y}\delta^{3}\left(z-y\right)\right)d^{3}z=\varepsilon^{-1}\left(q_{x}\phi_{y}\left(x\right)-q_{y}\phi_{x}\left(y\right)\right)$$In particular, if $q_x=q_y$ (as is apparently the case in the cited external reference), then the simple reciprocal relation $\phi_{y}\left(x\right)=\phi_{x}\left(y\right)$ indeed holds.

OK. But, besides the point charge, there will also be induced surface charge density on the conductor. So, the integration over $\rho$ should include integration over the surface of the conductor. However, you can show this integration equals zero using the fact that the conductor's surface is an equipotential surface and the net charge on the surface is zero.