Uniqueness theorem in uniform electric field

Question

Consider the following: An uncharged metal sphere of radius $R$ placed in a uniform electric field $\vec{E} = E_0 \hat{z}$. The field will push positive charge to the northern surface of the sphere, and symmetrically negative charge to the southern surface. This induced charge, in turn, distorts the field in the neighborhood of the sphere. Find the potential in the region outside the sphere.

The sphere is an equipotential we can set it to zero. Then by symmtery the entire $xy$ plane is at potential zero. Then $V$ does not go to zero but rather far from the sphere the field is $E_0 \hat{z}$ we thus have $$v \to -E_0z + C.$$

Since $V = 0$ in the equatorial plane, the constant $C$ must be zero. Then the boundary condition are $$V=0~~~\text{when }r=R \\ V \to - E_0 r \cos(\theta)~~~\text{for } r >> R .$$

Using the spherical form of Laplace's equation we get that the potential outside the sphere is $$V(r , \theta) = - E_0(r - \frac{R^3}{r^2})\cos(\theta).$$

Does the uniqueness theorem of Laplace's equation guarantee that this potential would be the same potential for say any uniform electric field $\vec{E}_0$ since the boundary conditions would be the same (except maybe requiring a coordinate tranformation) even though the direction of the uniform electric field might be different?

Answer 1

The entire x-y plane is obviously at the same potential since all the fields are strictly perpendicular to it (draw a diagram if you are confused). Since we choose the sphere to be at potential zero, the point on the sphere which cuts the x-y plane is also at zero potential, and hence the entire plane is at zero by the definition chosen w.r.t your uniqueness question, due to the spherical symmetry of the situation, a point fixed w.r.t the direction of $E_0$ shall always have the same potential

{Note: A point fixed in space will not have the same potential for different orientations of $E_0$.}