Two Wrongs Make Both Wrong, None Right

From the post "Maxwell, Planck And Particles",

 $\cfrac { 1 }{ \varepsilon _{ o }}\oint{ \cfrac { 1 }{ 4\pi a^2_{\psi} }}\cfrac{d\,q_{\psi}}{d\,t}\,d\,A_l= \oint_{2\pi a_{\psi}}{\cfrac{d\,(2 \pi mc)}{d\,t}}\,d\,\lambda= \oint_{2\pi a_{\psi}}{F_\psi}\,d\,\lambda$

If, we formulate instead,

$\oint { \cfrac { d\,  }{ d\, t } \left\{ \cfrac { q_{ \psi } }{ 4\pi \varepsilon _{ o }a^{ 2 }_{ \psi  } }  \right\}  } \, d\, A_{ l }=\oint _{ 2\pi a_{ \psi  } }{ \cfrac { d\, (2\pi mc) }{ d\, t }  } \, d\, \lambda \, =\oint _{ 2\pi a_{ \psi  } }{ F_{ \psi  } } \, d\, \lambda$

$E_{\psi}=\cfrac { q_{ \psi} }{ 4\pi \varepsilon _{ o }a^{ 2 }_{ \psi  } }$

$\cfrac { d\,  }{ d\, t } \oint { E_{\psi}  } \, d\, A_{ l }=\oint _{ 2\pi a_{ \psi  } }{ F_{ \psi  } } \, d\, \lambda$

This expression is without any appendages because if the $2\pi$ factor missing from the circular momentum term is the only reason

$\varepsilon _{ o }\ne\mu_o$,

then,

$\varepsilon _{ o }=\mu_o=\cfrac{1}{c}$  

after we made the correction, which would set,

$Z_o=1=\sqrt{\cfrac{\mu_o}{\varepsilon_o}}$

that space present equal resistance to $B$ and $E$ field.  With such a change to $\varepsilon_o$, the numerical value of $q$ changes also.

Let, consider again,

 $\cfrac { 1 }{ \varepsilon _{ o }}\oint{ \cfrac { 1 }{ 4\pi a^2_{\psi} }}\cfrac{d\,q_{\psi}}{d\,t}\,d\,A_l= \oint_{2\pi a_{\psi}}{\cfrac{d\,(2 \pi mc)}{d\,t}}\,d\,\lambda$

$\cfrac { 1 }{ 4\pi\varepsilon _{ o }}\oint{ \cfrac { 1 }{ 2\pi a^2_{\psi} }}\cfrac{d\,q_{\psi}}{d\,t}\,d\,A_l= \oint_{a_{\psi}}{\cfrac{d\,( mc)}{d\,t}}\,d\,\lambda= \oint_{2\pi a_{\psi}}{F_\psi}\,d\,\lambda$

without the correction to circular momentum.

Since,

$J_{\psi}=\cfrac { 1 }{ 2\pi a^2_{\psi} }\cfrac{d\,q_{\psi}}{d\,t}$

 $\cfrac { 1 }{ 4\pi\varepsilon _{ o }}\oint{ J_{\psi}}\,d\,A_l= \oint_{2\pi a_{\psi}}{F_\psi}\,d\,\lambda$

And then we make the adjustment to $\varepsilon _{ o }$, $4\pi\varepsilon _{ o }\rightarrow\varepsilon _{ new }$  to correct for the mssing $2\pi$ factor for circular momentum.

 $\cfrac { 1 }{\varepsilon _{ new }}\oint{ J_{\psi}}\,d\,A_l= \oint_{2\pi a_{\psi}}{F_\psi}\,d\,\lambda$

 Since, $\mu_{new}=\varepsilon_{new}=\cfrac{1}{c}$

 $\cfrac { 1 }{\mu_{ new }}\oint{ J_{\psi}}\,d\,A_l= \oint_{2\pi a_{\psi}}{F_\psi}\,d\,\lambda$

Why not simply make the adjustment $mc\rightarrow2\pi mc$, and obtain,

$\cfrac { 1 }{ 2\varepsilon _{ o }}\oint{ J_{\psi}}\,d\,A_l= \oint_{2\pi a_{\psi}}{F_\psi}\,d\,\lambda$

as in the post "Maxwell And Particles"???

This expression is wrong because it is still using the old definitions of $\mu_o$ and $\varepsilon_o$.