Fc=e2εo.14πr2
ac=Fce=eεo.14πr2
From the post "General Field Equation (factor 2 correction)", where the maximum ac is,
limr→0{eεo.14πr2}→c32
Consider,
e=es4πr2es
then, if we squeeze the electron to obtain the maximum acceleration as we would Earth to create a black hole,
limr→0{eεo.14πr2}=limr→0{esεo}=esεo
esεo≈c32
es≈c32εo
This would suggest charge is a surface phenomenon.
Otherwise,
e=ev.43πr3v
then
limr→0{eεo.14πr2}=limr→0{13evεorv}→0
However, by analogy this would also suggest that mass is a surface phenomenon too.
mss≈14πGc32
where mss is defined as
ms=mss4πr2ss
That gravitational effects of a mass shields the effects of the mass below it, when r→0, at the atomic level.
And so,
4πG.mss≈1εoes
Equivalence between mass and charge, at the atomic, r→0, surface density level.
My guess is,
mss≈es and that,
4πG≈1εo, as r→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.