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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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- Thread starter heman
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- #1

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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.??

- #2

quasar987

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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

[tex]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[/tex]

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

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Thanks Quasar

It's clear to me.

It's clear to me.

- #4

reilly

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Regards,

Reilly Atkinson

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