# Charge inside a metallic sphere

## Main Question or Discussion Point

Suppose we have charge inside a metallic spherical shell ,not in the middle,but near the boundary of the surface.The charge distribution near the charge will be dense but outside it will be uniform.
All this is clear to me that outside the metallic sphere will be equipotential,but what about the internal boundary??
Doesn't it have to satisfy the equipotentialness for the innermost boundary.??

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quasar987
Homework Helper
Gold Member
Suppose the "test charge" is located very near a certain area of the inner surface. Then the field produced by the test charge is stronger on that area of the conductor then it is on an area that is farther away from the test charge. Hence, the surface charge must concentrate more heavily on that part of the inner surface if it wants to cancel the field of the test charge. That's why the charges don't spread uniformly on the surface as soon as there is no symetry in the cavity and test charge(s).

In my opinion, the strange behavior is not from the charges on the inner surface, but from those on the outer surface. Anyway, it is not a trivial subject. Refer to Griffith's Electrodynamics 3rd edition pp.99 Exemple 2.6 and pp.118 for the complete explanation.

As for your actual question, Yes, that is the point of the non-uniform distribution. Here's a proof that the surface of a conductor is at equipotential with the rest of the conductor. Simple: chose your reference point in the calculation of the potential to be inside the conductor, and chose the path of integration to be inside the conductor as well. The only place where the field is non-zero is at the surface of the conductor itself, so your integral becomes

$$V(\vec{r}) = -\int_{\vec{\mathcal{O}}}^{\vec{r}}\vec{E}\cdot\ d\vec{r} = \lim_{max{||\vec{\Delta r}_i||}\rightarrow 0} \sum_{i=1}^n \vec{E}(\vec{r_i})\cdot \vec{\Delta r}_i = 0+0+...+\vec{E}(\vec{r_n})\cdot \vec{0} = 0+0+...+0=0$$

Hence the potential at the surface is 0, like at every point inside the conductor.

Thanks Quasar
It's clear to me.

reilly