Consider the electrostatic problem expressed by the Poisson equation $$\nabla^2\phi=-\rho$$ where the $\phi$ is the electric potential which vanishes at infinity and $\rho$ is the charge distribution that is compactly supported on the unit ball. The solution of this equation is $$\phi = G\rho$$ where $G$ is the Green's function operator.

Consider now the restricted operator $G$ mapping $\rho$ compactly supported on the unit ball to $\phi$ on the unit sphere $S:=\big\{x\in\mathbf R^3 \big|\,\|x\|=1\big\}$, or $$\phi\big|_S = G\rho.$$ There are infinitely many $\rho$'s corresponding to any given $\phi$. In fact, take any charge distribution function, reflect the charges with respect to the unit sphere. Compute the surface charge distribution thus generated on the unit sphere. The potentials of the surface charge combined is a null space charge distribution since the potential of each cancels the other in the exterior space of the ball. In other words the null space or kernel of the thus restricted $G$ is not empty. My question is how to characterize the null space or kernel of $G$, or all $\phi$ such that $G\rho=0$ on the unit sphere.

In more physical terms, I ask what nonzero charge distribution in a unit ball will result in zero electrical potential on the unit sphere.

  • $\begingroup$ Would the down-voter be brave enough to explain his rationale for downvoting this question and voice his objection in the open? $\endgroup$
    – Hans
    Commented Nov 23, 2021 at 3:42

1 Answer 1


If you are asking about the charge distribution inside the sphere, then any (suitably well behaved) interior charge distribution can be associated with a given potential $\phi(\vec{r})$ on the boundary. Moreover, there is nothing special about the spherical geometric in this case. The result is a general consequence of the most important uniqueness theorem in electrostatics: The potential in a bounded region of space $\Omega$ is uniquely determined by the combination of (1) the charged distribution inside the region $\Omega$—that is, $\rho(\vec{r})$ for $\vec{r}\in\Omega$—and (2) the potential on the boundary $\partial\Omega$—that is,$\phi(\vec{r})$ for $\vec{r}\in\partial\Omega$. The only additional conditions are regularity conditions on the prescribed $\rho|_{\Omega}$ and $\phi|_{\partial\Omega}$.

Your problem specifies only the value of the potential $\phi(\vec{r})$ on the boundary and asks what interior charge densities are consistent with this. So the answer is any $\rho$ that is sufficiently regular (regular enough that you can integrate over it, basically) will work. This actually makes a lot of physical sense, since you can wire up sectors of the surface $\partial\Omega$ to a bunch of large batteries with different voltages. This will set the voltage on each section of boundary to a specified value; with enough batteries and small enough sectors of $\partial\Omega$, you can essentially assign any voltage distribution on the surface—and this is true regardless of what additional sources are located inside!

On the other hand, this can seem confusing, as if the charge distribution in the interior has not made any difference in terms of determining $\phi$ on the surface. However, this is only because we have not said anything about the charge distribution outside the surface. The voltage on $\partial\Omega$ is determined by a combination of the charge inside and the charge outside. In the example in the previous paragraph, the charge outside took the form of charges on batteries and their attached conductors. The charge distribution outside $\Omega$ this is required to produce a given $\phi|_{\partial\Omega}$ can be enormous. For example, if (as in your original question), the region $\Omega$ is spherical, and it contains a total charge $Q_{{\rm in}}>0$, then in order to have the boundary be grounded, $\phi|_{\partial\Omega}=0$, the total negative charge outside must be greater in magnitude than the charge inside—that is, $Q_{{\rm out}}<-Q_{{\rm in}}$.

  • $\begingroup$ Let us continue this discussion in chat. $\endgroup$
    – Hans
    Commented Apr 28, 2021 at 18:21
  • 1
    $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. Note that chat is asynchronous: there is no requirement to participate in real time, any more than in comments. $\endgroup$
    – rob
    Commented May 1, 2021 at 18:01
  • $\begingroup$ Summarizing my valuation of this answer which was moved to chat, the answerer misunderstood the question and thus did not answer the question. Please refer to the question and chat for details. $\endgroup$
    – Hans
    Commented Nov 14, 2021 at 17:50

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