Suppose that the two conductors carry free charges $Q$ and $-Q$. Consider a closed surface enclosing the conductor with the charge $Q$ (but not the other conductor). We have, using Ohm's and Gauss' laws,

$$
I=\oint \mathbf{j} \cdot d \mathbf{S}=\oint \sigma \mathbf{E} \cdot d \mathbf{S}=\sigma \oint \mathbf{E} \cdot d \mathbf{S}=\sigma \frac{Q}{\varepsilon} .
$$

If the potential difference between the two conductors is $V$, we have $V=$ $I R=\frac{\sigma Q}{\varepsilon} R$, whence

$$
C=\frac{Q}{V}=\frac{\varepsilon}{\sigma R} .
$$

Numerically the capacitance between the conductors is

$$
C=\frac{80 \times 8.85 \times 10^{-12}}{10^{-4} \times 10^{5}}=7.08 \times 10^{-11} \mathrm{~F} .
$$
Two conductors are embedded in a material of conductivity $10^{-4} \Omega / \mathrm{m}$ and dielectric constant $\varepsilon=80 \varepsilon_{0}$. The resistance between the two conductors is measured to be $10^{5} \Omega$. Calculate the capacitance between the two conductors.