What fraction of a charge is induced on a surface via balayage? Consider a smooth, bounded domain $\Omega \subset \mathbb{R}^3$, and place a  charge $q>0$ at a point $z\in\mathbb{R}^3\setminus\overline\Omega$.  Via the concept of balayage, there is an induced surface charge density (i.e. signed Radon measure) $\nu$ supported on $\partial\Omega$ such that 
$$
\int \frac{d\nu(y)}{|x-y|} = \frac{q}{|x-z|}
$$ 
for all $x\in\Omega$.  Moreover, the total variation of $\nu$ is bounded by $q$. I think it is known that if $z$ lies in a bounded component of $\mathbb{R}^3\setminus\overline\Omega$, then this becomes an equality i.e. the total variation of $\nu$ equals $q$.   My question is what fraction of $q$ is the total variation of $\nu$ if $\overline\Omega$ does not fully enclose $z$?  Is it proportional to the solid angle subtended by $\overline\Omega$ from the perspective of point $z$ (i.e. the fraction of the view of infinity from $z$'s perspective that is blocked by $\overline\Omega$)?
 A: (Too long for a comment.)
If $\Omega = B(0, r)$ is a ball, then the fraction is given by $r/|z|$. More precisely, the harmonic reduction of $$u(x) = (4\pi)^{-1} q |x-z|^{-1}$$ in $\mathbb{R}^3 \setminus \Omega$ is given by $$v(x) = (4\pi)^{-1} q r |z|^{-1} |x-z^*|^{-1}$$ for $x \in \mathbb{R}^3 \setminus \overline{\Omega}$, where $$z^* = r^2 |z|^{-2} z$$ is the image of $z$ under inversion with respect to the sphere $\partial B(0, r)$. Of course, $v(x) = u(x)$ in $\Omega$. The expression for $v$ follows by a standard calculation related to the Kelvin transformation in classical potential theory: one needs to verify that $v(x) = u(x)$ for $x \in \partial \Omega$ (a lengthy calculation), that $v$ is harmonic in the complement of $\Omega$, and that it decays at infinity. Since in the complement of $\Omega$ the function $v$ is simply the potential of a point mass $q r |z|^{-1}$ at $z^*$, it follows that $v$ is the potential of a measure with total mass $q r |z|^{-1}$, as desired.
A: Balayage is a positive operator so the measure $\nu$ obtained by sweeping out the positive charge $q\delta_{z}$, onto $\partial\Omega$, is a positive measure as well. The mass of $\nu$ can be expressed in a nice way if we appeal to the (newtonian) capacity $\text{cap}(\overline\Omega)$, the equilibrium measure $\omega_{\overline\Omega}$ of $\overline\Omega$, and its potential $U^{\omega_{\overline\Omega}}$. Indeed,
\begin{align*}
\|\nu\| & =\int d\nu= \text{cap}(\overline\Omega)\int U^{\omega_{\overline\Omega}}d\nu =
\text{cap}(\overline\Omega)\int U^{\nu}d\omega_{\overline\Omega}
\\[10pt]
& =
\text{cap}(\overline\Omega)\int U^{q\delta_{z}}d\omega_{\overline\Omega}= q~\text{cap}(\overline\Omega)\int U^{\omega_{\overline\Omega}}d\delta_{z}
=q~\text{cap}(\overline\Omega)U^{\omega_{\overline\Omega}}(z).
\end{align*}
Such results can be found in the book by Landkof, Foundations of modern potential theory, Chapter IV.
In the case where $\Omega$ is the ball of radius $r$, considered by Mateusz Kwasnicki, one has
$\text{cap}(\overline\Omega)=r$ and $U^{\omega_{\overline\Omega}}(z)=U^{\delta_{0}}(z)=|z|^{-1}$, and thus we recover that the fraction $\|\nu\|/q$ equals $r|z|^{-1}$.
