May The Hunt Begin

If,

$q=m.4\pi a_{\psi}^{ 2 }$

from the post "Pound To Rescue Permittivity" dated 30 May 2016, also,

$\cfrac{q}{\varepsilon_o}=4\pi a^2_{\psi}m.2c^2ln(cosh(\pi))$

$\varepsilon_o$ resists the spread of $q$ into the region with permittivity, $\varepsilon_o$.

If each charge is a portal to a storage of flux, $\varepsilon_o$ resist or aid ($\varepsilon_o\lt 1$) the flow of flux from the storage.

$\cfrac{q}{\varepsilon_o}=q.2c^2ln(cosh(\pi))$

$\cfrac{1}{\varepsilon_o.2ln(cosh(\pi))}=c^2$

And we start a treasure hunt!

Note:  This posts has lost its original meaning.  What was it about?  $ln(cosh(\pi))$ is arbitrary.  For a basic particle this could be,

$ln(cosh(\cfrac { G }{ \sqrt { 2{ mc^{ 2 } } }  }a_{\psi\,c}))=\cfrac{1}{4}$

$ln(cosh(0.7369))=\cfrac{1}{4}$