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I have read that the inverse square potential outside a spherical shell is the same as that due to a point mass/charge at the origin of the same mass/charge, and that in general, for the Yukawa potential $\exp(-ar)\over r$ where $a$ is a constant and $r$ the radial distance, the outside potential due to a spherical shell is the same as that due to a point charge of mass $M\sinh(aR)\over aR$ where $R$ is he radius of the shell.

How can I prove the latter properly? I can see that the former is a special case of the latter, so proving the latter would suffice.

Thanks!

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The Yukawa potential $u$ in empty space satisfies $\Delta u = a^2 u$. There are two linearly indpendent spherically symmetric solutions of this differential equation: $\exp(-ar)/r$ and $\exp(+ar)/r$. The potential outside the spherical shell should be spherically symmetric and decay, not grow, as $r \to \infty$, so this must be $C \exp(-ar)/r$ for some constant $C$. This is the potential of a point mass $C$ at the origin. On the other hand, inside the shell the potential should be nonsingular at the origin, so it is of the form $A \sinh(ar)/r$, which has a finite limit $Aa$ as $r \to 0+$. For the potential to be continuous at the shell itself (radius $R$), we need $A \sinh(aR)/R = C \exp(-aR)/R$, i.e. $C = A \sinh(aR) \exp(aR)$.

Now if the shell has radius $R$, the origin is at distance $R$ from all points of the shell, so the potential at the origin is the same as it would be if all the mass of the shell was gathered into one point, still at distance $R$, namely $M \exp(-aR)/R$ where $M$ is the mass of the shell. Thus $Aa = M \exp(-aR)/R$. Thus we get $C = M \sinh(aR)/(aR)$.

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