The Price Of Three

...an idiot because,

$T_k=r*\cfrac{T}{Tg}$

where $r=\cfrac{q_g}{\varepsilon_g k \Delta T}$

when

$T_g=\cfrac{\rho_g}{\varepsilon_g}=\cfrac{\rho_c}{\varepsilon_c}=T$

as measured on the inner wall of the containment.

$\cfrac{T}{T_g}=1$

and so,

$T_k=r$

but,

$\Delta T=T_g-T=0$

and so,

$T_k=\cfrac{q_g}{\varepsilon_g k \Delta T}\rightarrow \infty$

What went wrong?

$T=\cfrac{\rho_g}{\varepsilon_g}=\cfrac{\rho_c}{\varepsilon_c}$

but

$\cfrac{\rho_c}{\varepsilon_c}\ne\cfrac{\rho_{\small{g,\,P\rightarrow \infty}}}{\varepsilon_g}$

in the volume $V_m$, the temperature charge did not all move to the surface.  If they did, the surface density will be numerically the same as volume density.  What then is the point of $V_m$ or $r=3$?

$T=\cfrac{\rho_c}{\varepsilon_c}=\cfrac{\rho_g}{\varepsilon_g}=\cfrac{rho_v}{\varepsilon_g}$

as $P\rightarrow \infty$

such that $T$ now measures the temperature of the volume $V_m$ with a certain volume temperature charge density and not just its surface temperature charge.

It seems that pressure must still be high, and we can be off by a factor as much as $3$.  The change in size $V\rightarrow V_m$ simply removes the factor $\small{\cfrac{1}{3}}$ from the expression,

$r=\cfrac{q_g}{3\varepsilon_g k \Delta T}$

to obtain,

$r=\cfrac{q_g}{\varepsilon_g k \Delta T}$

the assumption of high pressure remains.  The price of this $3$ is a big volume.

Have a nice day.