Uniqueness theorem in uniform electric field

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?

• Why is the entire x-y plane at potential zero? – GeeJay Aug 23 '16 at 14:06
• @JayJay Good question...I'm trying to figure that out as well... This is an example from Griffiths "Introduction to electrodynamics" fourth edition, page 145 Example 3.8. if you are interested. Let me know if you make any progress... – Alex Aug 23 '16 at 15:20

• @Lelouch Hi, can the reason be seen from the fact that $V = -\int_{a}^{b} \vec{E} \cdot dI$, since in the $xy$ plane the $dI$ is perpendicular to $\vec{E}$, we would get that $V$ is zero everywhere on $xy$ plane if $a$ is taken as a point on the surface of the sphere and in the $xy$ plane? Then secondly, why is it okay to just set the potential of sphere to zero if it is in equipotential as a conductor? Will the potential at points around the sphere then also be altered? Thanks. – user100411 Sep 21 '16 at 16:54