Opps! Lucky Me

And we have a problem,

$F=\cfrac{1}{12\pi r}\cfrac{d\,\psi}{d\,r}$

$RHS=m^{-1}.Jm^{-3}m^{-1}$

$J=kgm^2s^{-2}$

$RHS=kgm^{-3}s^{-2}$

$LHS=kgms^{-2}$

$m^{-4}$ missing?

WTF!  This also means that the post "Flux It" dated 21 Nov 14 is also wrong.

But $\varepsilon_o$ to the rescue!  It is possible to introduce a constant $\varepsilon_o$ to remove the inconsistency in unit dimensions and at the same time introduce the unit for coulomb charge $C$.  But it is first, more important to realize that $\psi$ due to a point source is distributed uniformly in a sphere with the point source as center.  The force due a change in $\psi$ along a radial line, at $r$ distance from the center, is also divided evenly over the surface area of the sphere $4\pi r^2$.  As such the force at $r$, along a radial line, is the force due to $\psi$ per unit surface area of the sphere of radius $r$.  In this way, strictly speaking,

$LHS.m^{-2}=RHS$

$LHS.m^{-2}=kgms^{-2}m^{-2}=kgm^{-1}s^{-2}$

still $m^{-2}$ missing!

What happened?  Then comes $\varepsilon_o$ with units of $C^2N^{-1}m^{-2}$. Since $\varepsilon_o$ would be in the denominator on the RHS, we have $N.C^{-2}m^{2}$.  This is not satisfactory.

Maybe the second division by $4\pi r^2$ in the derivation for $F$ along a radial line is not required?

We could have, energy density over a surface area of $4\pi r^2$, $\psi_A$ as

$\psi_A=\psi.4\pi r^2$

If $\Delta E$ is the energy in the thickness of the thin spherical shell around the center particle of radius $r$,

$\Delta E=\psi_A.\Delta r$

When this shell is infintely thin,  $\Delta r\to 0$,

$\cfrac{d\,E}{d\,r}=\lim\limits_{\Delta r\to0}{\cfrac{\Delta E}{\Delta r}}=\psi_A$

From which we obtain,

$F_s=-\cfrac{d\,E}{d\,r}=-\psi_A$

Strictly speaking we have lost the direction of $F_s$ when we substituted in $-\psi_A$ because
$\psi_A$ cannot be negative,

$|F_s|=\psi.4\pi r^2$

But this force is distributed over the surface of the sphere of radius $r$, as such the force along a radial line is,

$|F|=\cfrac{|F_s|}{Area}=\cfrac{1}{4\pi r^2}\psi.4\pi r^2$

$|F|=\psi$

If $F$ is to have a inverse square law dependence then,

$\psi=\cfrac{D}{r^2}$

where $D$ is a constant.  Using $E=mc^2$ we have

$m=\cfrac{D}{c^2r^2}$

but $m$, mass density is in the time dimension so we have instead,

$m=\cfrac{D}{c^2t^2}$

which then suggests that $m$ stretches out in time in a corresponding way as $\psi$ stretches out in space.  And instead of just a equivalence relationship we have a transform,

$E(r)=m(t)c^2$

from space ($r$) to time ($t$).

Simplicity rides again.