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We know that potential of a conductor is increased when charge is applied to it. For example a neutral body has charge $0$. When we apply positive charge to it it's potential increases to $V$ since we now have to do positive work to bring a unit positive charge from infinity since the positive charge of the conductor will repel that unit positive charge. But we also know that potential at a distance $r$ from the charge is $\frac{kq}{r}$. Here $r$ is $0$ since we are trying to find the potential at that point. Doesn't it mean potential of the conductor is infinite?

madness
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  • When you have a finite object you don't have $\frac{1}{r}$ diverging. You just need to match the field outside to the potential on the surface of the object. – Dan Jan 13 '22 at 05:08

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The potential outside a charge distribution is $\frac{kq}{r}$ only if the total charge ($q$) is distributed with spherical symmetry (for example, the conductor is a spherical shell). Here, $r$ is the distance from the center of the conductor, so if the conductor has radius $R$, then the potential at the surface of the conductor is $kq/R$ (not infinite).

Inside the spherical conductor, the potential is no longer $\frac{kq}{r}$. That's why the potential at the center of the conductor doesn't go to infinity.

The correct expression for the potential, both inside and outside, can be calculated using Gauss' Law (again, assuming there's spherical symmetry). Further explanations and graphs can be found here: http://hyperphysics.phy-astr.gsu.edu/hbase/electric/potsph.html

dkarkada
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