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The situation is illustrated in the attachment. I try to find the force on q by the method of images.

I know that the potential is 0 at x=R and at infinity, and I know that The potential obeys a certain Poisson equation between those points. I will try to put a charge q' somewhere at x<R that makes it so V=0 at x=R and at infinity. That way, according to the uniqueness theorem, the potential between x=R and infinity created by the 'sphere-q' system will be the same as the potential created by the 'q'-q' system.

My intuition tells me that I can pretty much put q' anywhere I want and I'll be able to create V=0 at x=R just by adjusting the magnitude of the charge. So I chose to put it at x=0. The potential of this charge configuration is

[tex]V(x) = \frac{1}{4\pi \epsilon_0}\left(\frac{q}{|x-a|}+\frac{q'}{x}\right)[/tex]

and we have V(R) = 0 iff

[tex]q' = -\frac{R}{|R-a|}q = -\frac{R}{a-R}q[/tex]

So the force on q is just that exerted by q' at this particular location, i.e.

[tex]F_{\rightarrow q}=-\frac{1}{4\pi\epsilon_0}\frac{Rq^2}{(a-R)(a)^2}[/tex]

Yes? Or do I have the method of images all wrong?

I know that the potential is 0 at x=R and at infinity, and I know that The potential obeys a certain Poisson equation between those points. I will try to put a charge q' somewhere at x<R that makes it so V=0 at x=R and at infinity. That way, according to the uniqueness theorem, the potential between x=R and infinity created by the 'sphere-q' system will be the same as the potential created by the 'q'-q' system.

My intuition tells me that I can pretty much put q' anywhere I want and I'll be able to create V=0 at x=R just by adjusting the magnitude of the charge. So I chose to put it at x=0. The potential of this charge configuration is

[tex]V(x) = \frac{1}{4\pi \epsilon_0}\left(\frac{q}{|x-a|}+\frac{q'}{x}\right)[/tex]

and we have V(R) = 0 iff

[tex]q' = -\frac{R}{|R-a|}q = -\frac{R}{a-R}q[/tex]

So the force on q is just that exerted by q' at this particular location, i.e.

[tex]F_{\rightarrow q}=-\frac{1}{4\pi\epsilon_0}\frac{Rq^2}{(a-R)(a)^2}[/tex]

Yes? Or do I have the method of images all wrong?

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