Does Size Matter?

Does size matter?

See this quantum charge being $\frac{1}{3}e$.  I think it is $\frac{1}{\pi}e$.   Without the right charge, then the dipole is wrong... What the ${F}_{n}$, $F$?  What is the nature of electrostatic forces?  What if space is not is electrically neutral?

Is there a force on the dipole simply because physically the negative charge is smaller and the field lines are denser where they terminates on its surface?  The field line tends to space apart so there is a net force rightwards?

If space is not neutral but has a net negative charge, then dipole as a whole will experience a net force due the physical shape of the dipole.  Consider a cone with open radius  $R_1$  and  $R_2$ along $R$, an elemental surface $ds$ on the cove surface is given by

$ds = dp.dh$

$dp = R.d\phi$,   $dh= \cfrac{{dx}}{\cos{(\theta)}}$

$ds =  R.d\phi.\cfrac{dx}{\cos{(\theta)}}$

The elemental charge on the surface is,

$dq=\sigma.ds$     where    $\sigma$ is the surface charge density of space around the cone.  The potential at the apex is

$dV= \cfrac{1}{4\pi\varepsilon_o}\cfrac{\sigma.ds}{r}$

Since,

$r=\cfrac{x}{\cos{(\theta)}}$     $R=x.\tan{(\theta)}$

$dV=\cfrac{1}{4\pi\varepsilon_o}\sigma \tan{(\theta)}d\phi dx$

$x$ cancels away.

$V = \cfrac{\sigma}{4\pi\varepsilon_o}.\tan{(\theta)}\int^{2\pi}_{0}\int^{R_2/\tan{(\theta)}}_{R_1/\tan{(\theta)}}{dx}$

$V = \cfrac{\sigma}{2\varepsilon_o}.\tan{(\theta)}\int^{R_2}_{R_1}{dx}\cfrac{1}{\tan{(\theta)}}$

$E = -\cfrac{dV}{dx}= -\cfrac{\sigma}{2\varepsilon_o}$

There is net force on the structure through the apex along the $x$ direction.  This might be the perpetual force on the dipole that gives it light speed.  Direction, but no magnitude to be conclusive.