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 $$ \int \frac{d\nu(y)}{xy} = \frac{q}{xz} $$ 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 $\overline\Omega$)?

1$\begingroup$ This seems to be the probability that a 3D Brownian motion started at $z$ ever hits $\overline{\Omega}$. It is not possible to express this quantity in a closedform way for general $\Omega$. $\endgroup$ – Mateusz Kwaśnicki Jul 3 at 6:26

1$\begingroup$ $q$ is missing in your displayed formula. $\endgroup$ – Alexandre Eremenko Jul 3 at 14:40

$\begingroup$ @AlexandreEremenko fixed, thank you $\endgroup$ – Ben Ciotti Jul 3 at 19:37

$\begingroup$ @MateuszKwaśnicki ah I didn't realize that, thank you for the insight. I wonder if there is a shape such that the total variation of the induced surface charge is some prescribed fraction of q? $\endgroup$ – Ben Ciotti Jul 3 at 19:40
(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 $$u(x) = (4\pi)^{1} q xz^{1}$$ in $\mathbb{R}^3 \setminus \Omega$ is given by $$v(x) = (4\pi)^{1} q r z^{1} xz^*^{1}$$ 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.

$\begingroup$ Very interesting, I see my solid angle supposition considerably underestimated the induced charge; this makes much more sense in light of the brownian motion comment, thank you $\endgroup$ – Ben Ciotti 2 days ago