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An uncharged metal sphere of radius $R$ is placed in a uniform electric field $\vec{E} = E_0 \, \hat{z}$. The field will push the positive charges to the "northern" surface of the sphere, and the negative charges to the "southern" surface.

Griffiths in Example $3.8$ says that the sphere is an equipotential. But how do I know that?

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    $\begingroup$ A conductor is equipotential because the E field inside it is zero. $\endgroup$ – velut luna Apr 18 '16 at 1:41
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The electric field induces some charge on the sphere. This charge on the sphere creates it's own electric field. The metal sphere will have a constant electric field throughout the surface as you said that the applied field is uniform. There is no charge inside. So there can't be any electric field inside the sphere. Otherwise it will violate Gauss's law (imagine that on applying Gauss's law to the sphere, you have no charge inside the volume, but still you get electric field diverging from the surface). The charges reside only on the surface and we found that the electric field at the surface is constant as the radius of the sphere is constant. We have the electric field is the gradient of scalar potential V. So no electric field inside the sphere means that the gradient of potential inside the sphere is zero. This means that the potential inside the sphere is a constant. This is why the sphere is said to be equipotential. The potential starts to change at the point when we reach the surface of the sphere.

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A metal sphere means that it is a conducting sphere. Any difference of the potential would cause an electric current to flow until this difference is balanced, which means that no difference in potential can remain thus equipotential.

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