Optimal position of negative charge to "neutralise" a positive distribution

Question

This question stems from trying to understand the notion of center of charge and if the analytical definition of this center depends on what exactly is minimized (the dipole moment or the total potential energy?). Which are the useful properties of the "center of charge" and how is it obtained in terms of a minimization procedure?

Optimisation problem: Imagine we have a distribution of positive charge $\rho(\mathbf{x})$ such that its support (the region where it is non-zero) is compact. Also, if we integrate $\rho(\mathbf{x})$ over the whole space we have a finite total charge $Q>0$. We now add a single point charge $-Q$. Where do we have to put it in order to minimise the total electric field?

Example (the trivial case): Clearly, if $\rho(\mathbf{x})=Q\delta(\mathbf{x-y})$, then the optimum is to put the negative charge in $\mathbf y$: in this case we can completely kill the total electric field everywhere: any other position will give rise to a dipole.

The general case: My feeling is that we have to consider the total potential energy of the resulting charge distribution and minimise it with respect to the position of the negative point charge... or do we have to try to minimise the dipole moment? The procedure should return the position of the "center of charge" of $\rho$. Is this always the case regardless of whether we try to minimise the total potential energy or the dipole moment?

Answer 1

Assuming that this "neutralization" corresponds to a minimization of the action upon a test charge, and assuming further that the problem is three dimensional, the procedure would be to minimize the resulting electric potential w.r.t the position of the negative point charge. As you still solve Poisson's equation, the outline would be as follows:

Divide the charge distribution in two parts

The first part is the arbitrary configuration $\sigma(\mathbf{x}) $ whereas the second part is just a delta at some position $\delta(\mathbf{x} - \mathbf{w}) $ For this configuration you would calculate the electric potential by using Green's functions. Now the Green's function in 3d is simply $$G(\mathbf{x} - \mathbf{y}) =\frac{1} {\left|\mathbf{x} - \mathbf{y} \right|} $$ which in turn leads to the electric potential

$$\phi_\sigma(\mathbf{x} ; \mathbf{w})= \int \frac{\sigma(\mathbf{y} )-Q\delta (\mathbf{y}-\mathbf{w} ) } {\left| \mathbf{x} - \mathbf{y} \right|} \mathrm{d}\mathbf{y} $$

The minimization procedure

One way to go about this was to exchange the delta distribution with one of its limiting sequences $$ \delta(\mathbf{y}-\mathbf{w})= {\lim_{\small{n\rightarrow \infty}} } f_n(\mathbf{x} - \mathbf{y}) $$ and to prove that minimization w.r.t $\mathbf{y} $ is mathematically sound, for example by checking that the sequence converges uniformly (maybe at least in a weak sense) and thus asserting that at least in a weak sense variational calculus can be employed.

Assuming, this validity of convergence has been shown, minimizing the potential $(\phi_\sigma \mathbf{x} ;\mathbf{w}) $ w.r.t $\mathbf{w} $ leads to

$$ \nabla_{\mathbf{w}} \phi_\sigma (\mathbf{x} ;\mathbf{w})=0 \\ \lim_{\small{n \rightarrow \infty}} \nabla_\mathbf{w} \int\frac{\sigma(\mathbf{y}) - Qf_n(\mathbf{y} - \mathbf{w}) } {\left|\mathbf{y} - \mathbf{x} \right|} \mathrm{d} \mathbf{y} =0 \\ \mathrm{Hess}(\phi_\sigma(\mathbf{x} ;\mathbf{w}) ;\mathbf{w})\ \ \ \mathrm{\ positive\ definite}$$

The first and secondequation provides a way to find position where the electric potential becomes extreme and the third equation calculating the Hessian of the potential distinguishes between maxima, minima and saddle points.

It is not a priori clear that this way leads to results but there might me certain families of charge distributions which allow for this.