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I'm studying the Method of Images and I seemed to have come to a conundrum. Method of Images takes advantage of grounded objects, (I am currently studying spheres), to set boundary conditions. However, how would one use the idea of MoI to set the potential of a conducting sphere as constant?

Since $E = \nabla V$, a constant potential would be the electric field inside would be constant? Therefore, a density of charge inside the sphere?

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I edited your question, to fix $\nabla E = V$, which should be $E=\nabla V$. but perhaps that typo is the source of your equation. –  askewchan Feb 26 '13 at 2:04
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Also note that conundrum is a noun and not a verb. –  Emilio Pisanty Feb 26 '13 at 16:20
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Regardless of whether it is grounded, if there were a nonzero electric field inside a conductor, it would push the charge carriers around until there were no longer forces on them.

Thus: A perfect conductor, grounded or isolated, will have a surface (and volume) of equal potential, and the electric field inside will be zero.

If the conductor is not grounded, then there will be some net charge (possibly zero) on the conductor. If it's a symmetric shape (a sphere) then it will have a uniform surface density, namely the total charge over the total surface area ($Q / 4\pi r^2$).

You cannot assume to know $Q$ on the sphere. Whether or not $Q=0$, there will be $E=0$ inside the sphere (Gauss Law), and outside the sphere there will be $E\neq 0$ if $Q\neq0$.

Adding an image charge at the origin would actually create a nonzero $E$ inside the sphere, which you cannot have inside a conductor (I'm assuming the sphere is solid). The charge on the surface of the sphere already cancels its own $E$ field inside the sphere. If you want to cancel the $E$ field outside the sphere, then an image at the origin would do that.

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So if I were to find the potential outside of a conducting sphere that is not grounded, the potential inside would have to be constant. Therefore that second image charge would have to be placed at the origin to mimic if it were grounded. –  julesverne Feb 26 '13 at 17:53
    
@julesverne you really don't need an image charge to solve this, assuming you're trying to find the potential everywhere, just use Gauss' law. –  askewchan Feb 26 '13 at 17:56
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