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Example 3.8: An uncharged metal sphere of radius R is placed in an otherwise uniform electric field $\vec{E} = E_o \hat{z}$. The field will push the positive charge to the northern surface of sphere, and-symmetrically- negative charge to the southern surface (Fig 3.24). This induced charge, in turn, distorts the field in the neighbourhood of the sphere. Find the potential in the region outside the sphere enter image description here

Point of doubt in solution: Sphere is taken as zero potential. Then by symmetry the entire xy plane is at zero potential.

This is a question from Griffiths Introduction to electrodynamics, the details of which have been mentioned in the figure itself. What I am not being able to understand is, why should the potential on the xy plane be zero?

If I walk radially outward on the xy plane from the surface of the sphere which has zero potential, am I never going to encounter a field vector which has a radial component?

What I have in my mind is that, the surface of the sphere being equipotential, the field on the xy plane very close to the surface would be radial, hence the potential on the xy plane isn't constant. How does one then justify the xy plane to have a zero potential?

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The electric field is equal to the external field plus the field generated by the charges on the sphere via Coulomb's law.

In the $xy$ plane, the external field doesn't have any radial component, and the radial component of the field from every positive charge on the sphere is cancelled by the corresponding negative charge on the other side. Thus the electric field has no radial component and the whole plane is has the same potential.

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  • $\begingroup$ Do you have a proof of your statement $\endgroup$ May 29, 2021 at 14:46
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The electric field induced in the sphere is equal and opposite to the external electric field for a net field within the sphere of zero. If there was a field in the sphere the charges in the sphere would redistribute themselves until the field vanished.

Hope this helps

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  • $\begingroup$ Nope. I was talking about whether the potential on the xy plane would be equal to that of the sphere. Maybe at any point on the xy plane, outside the sphere, the electric field—the external plus that due to the ball, would entirely be along the z axis, so there wouldn't a radial component of electric field that would cause the potential on the xy plane to change. $\endgroup$
    – Anu3082
    Feb 22, 2020 at 12:44

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