# How to make a nongrounded conductor have equipotential?

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 at 2:04
Also note that conundrum is a noun and not a verb. – Emilio Pisanty Feb 26 at 16:20

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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 thank you for your edit. It was a mistype when I was typing the question. I understand that a perfect conductor would have a surface of equal potential, but what if I want to make a non grounded sphere have equi potential inside? Method of Images tells me that if I introduce a charge to correct boundary conditions, it will give me that specific case. I can't seem to figure out where though.. – julesverne Feb 26 at 16:00 @julesverne see my edit – askewchan Feb 26 at 16:17 Can I interpret the conductor as "neutrally charged"? If there was a net charge in the sphere, there would be an electric field. Therefore, the second charge that must be placed to create an equipotential throughout the non grounded sphere, would be at the origin, and would have a charge of -Q? As that would cancel out the charge on the surface of the sphere, creating no electric field, giving a constant potential. – julesverne Feb 26 at 17:12 @julesverne: I edited the answer again. – askewchan Feb 26 at 17:18 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 at 17:53
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