Fc=e2εo.14πr2
ac=Fce=eεo.14πr2
From the post "General Field Equation (factor 2 correction)", where the maximum ac is,
lim
Consider,
e=e_{s}4\pi r_{es}^2
then, if we squeeze the electron to obtain the maximum acceleration as we would Earth to create a black hole,
\begin{matrix} \lim \\ r\rightarrow 0 \end{matrix}\left\{\cfrac{e}{\varepsilon_o}.\cfrac{1}{4\pi r^2}\right\}=\begin{matrix} \lim \\ r\rightarrow 0 \end{matrix}\left\{\cfrac{e_{s}}{\varepsilon_o}\right\}=\cfrac{e_{s}}{\varepsilon_o}
\cfrac{e_s}{\varepsilon_o}\approx c^\cfrac{3}{2}
e_s\approx c^\cfrac{3}{2}\varepsilon_o
This would suggest charge is a surface phenomenon.
Otherwise,
e=e_v.\cfrac{4}{3}\pi r_{v}^3
then
\begin{matrix} \lim \\ r\rightarrow 0 \end{matrix}\left\{\cfrac{e}{\varepsilon_o}.\cfrac{1}{4\pi r^2}\right\}=\begin{matrix} \lim \\ r\rightarrow 0 \end{matrix}\left\{\cfrac{1}{3}\cfrac{e_{v}}{\varepsilon_o}r_{v}\right\}\rightarrow 0
However, by analogy this would also suggest that mass is a surface phenomenon too.
m_{ss}\approx \cfrac{1}{4\pi G}c^{\cfrac{3}{2}}
where m_{ss} is defined as
m_{s}=m_{ss}4\pi r_{ss}^2
That gravitational effects of a mass shields the effects of the mass below it, when r\rightarrow 0, at the atomic level.
And so,
{4\pi G}.m_{ss}\approx \cfrac{1}\varepsilon_o e_s
Equivalence between mass and charge, at the atomic, r\rightarrow0, surface density level.
My guess is,
m_{ss}\approx e_s and that,
{4\pi G}\approx \cfrac{1}\varepsilon_o, as r\rightarrow 0
The question of how much for points is answered by volume density. The same question for lines is answered by area density, for an area defined perpendicular to the lines. Lines in a volume density, has redundant length information and is blind to changing area densities within the volume.
Have a nice day.