Consider an arrangement of two isolated conducting surfaces of any shape that form a capacitor. Use the definitions of capacitance (Eq. (2) in this chapter) and resistance (Eq. (14) in Chapter 5) to show that when the region between the conductors is filled with either conductive material (conductivity \(\sigma\) ) or a perfect dielectric (permittivity \(\epsilon\) ), the resulting resistance and capacitance of the structures are related through the simple formula \(R C=\epsilon / \sigma .\) What basic properties must be true about both the dielectric and the conducting medium for this condition to hold for certain?

The capacitance of the capacitor is given by the equation: $$C = \dfrac{Q}{V}$$ Where \(C\) is the capacitance, \(Q\) is the charge on the plates, and \(V\) is the potential difference between the plates. The resistance of the conductor can be defined as follows: $$R = \dfrac{V}{I}$$ Where \(R\) is the resistance and \(I\) is the current flowing through the conductor.

In case of a conductive material with conductivity \(\sigma\), the current density \(J\) is related to the electric field \(E\) as: $$J = \sigma E$$ For a dielectric material with permittivity \(\epsilon\), the displacement field \(D\) is related to the electric field \(E\) as: $$D = \epsilon E$$

The electric field \(E\) in both cases can be related to the potential difference \(V\) between the plates as: $$E = \dfrac{V}{d}$$ Where \(d\) is the distance between the plates. Using this relation, we can rewrite the capacitance and resistance formulas as follows. For capacitance: $$C = \dfrac{Q}{V} = \dfrac{Qd}{Vd} = \dfrac{Q}{ED} = \dfrac{Q}{\epsilon E^2 d}$$ For resistance: $$R = \dfrac{V}{I} = \dfrac{Ed}{I} = \dfrac{Ed}{Jd} = \dfrac{E}{J} = \dfrac{E}{\sigma E}$$

Multiplying the capacitance and resistance equations, we get: $$RC = \dfrac{Q}{\epsilon E^2 d} \times \dfrac{E}{\sigma E} = \dfrac{Q}{E} \times \dfrac{E^2}{\sigma \epsilon}$$ Notice that the charge \(Q\) is related to the current \(I\) as \(Q = I t\), where \(t\) is the time of charging. Therefore, we can substitute \(Q = It\) into the equation: $$RC = \dfrac{I t}{\epsilon E^2 d} \times \dfrac{E^2}{\sigma \epsilon}$$ The electric field terms, \(E^2\), cancel each other out, and we're left with the equation: $$RC = \dfrac{\epsilon}{\sigma} \times \dfrac{I t d}{Id}$$ The current \(I\) and distance between plates \(d\) also cancel each other, leaving us with the desired result: $$RC = \dfrac{\epsilon}{\sigma}$$

For the derived formula \(RC = \epsilon/\sigma\) to hold true, the following basic properties must be satisfied for both dielectric and conducting medium: 1. The material must be homogeneous and isotropic, meaning it has uniform properties in all directions. 2. The permittivity \(\epsilon\) and conductivity \(\sigma\) should be constant throughout the material. 3. The electric field and current density must be uniform between the conducting plates. With these conditions satisfied, the derived formula \(RC = \epsilon/\sigma\) will hold true for the given capacitor system consisting of two conducting surfaces separated by either a conductive material or a perfect dielectric.