# Potential in the x-y plane for an uncharged metal sphere kept in an external field pointing in the $\hat{z}$ direction

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

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?

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.