...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.
