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?

  • 2
    $\begingroup$ Why is the entire x-y plane at potential zero? $\endgroup$
    – GeeJay
    Commented Aug 23, 2016 at 14:06
  • $\begingroup$ @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... $\endgroup$
    – Alex
    Commented Aug 23, 2016 at 15:20

1 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 $E_0$.}

  • $\begingroup$ Thanks for your response. Just to confirm the last part of your answer which addresses the actual question of the post: Are you saying that the symmetry of the given problem allows me to use the uniqueness theorem the way I did, but generally this is not true? $\endgroup$
    – Alex
    Commented Aug 23, 2016 at 15:49
  • $\begingroup$ For you to apply uniqueness, the boundary values of the potential must be specified. Although it is specified at the surface of the conductor, it is not specified at infinity. Thus you cannot apply the uniqueness here $\endgroup$
    – Lelouch
    Commented Aug 23, 2016 at 16:53
  • $\begingroup$ Im merely saying that the potential will be same w.r.t the orientation of the E field due to spherical symmetry. $\endgroup$
    – Lelouch
    Commented Aug 23, 2016 at 16:54
  • $\begingroup$ @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. $\endgroup$
    – user100411
    Commented Sep 21, 2016 at 16:54
  • $\begingroup$ It is okay to define the potential as zero in this case, for which reason, it will no longer be zero at infinity as we encounter generally. This is merely a valid choice to simplify calculations, at the sacrifice of making the potential non zero at infinity , while zero on the plane and the sphere. $\endgroup$
    – Lelouch
    Commented Sep 21, 2016 at 17:01

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