# Using graphite or graphene as two parallel plates in the capacitance experiment

We usually use copper as the parallel plates in the parallel plates capacitance experiment. I'm wondering would it makes any differents on capacitance. If we substitute copper with graphite or graphene.

By Gauss' Law $$\int EdA=\frac{q}{\epsilon_0},$$ for E is constant at all points on Gauss' surface $$\Rightarrow E\int dA=\frac{Q}{\epsilon_0} \\\Rightarrow E\times A=\frac{Q}{\epsilon_0} \\\Rightarrow E=\frac{Q}{\epsilon_0 A} \\\Rightarrow Ed=V=\frac{Qd}{\epsilon_0 A}$$

$$C\equiv \frac{Q}{V} \\\Rightarrow C=\frac{Q}{\frac{Qd}{\epsilon_0 A}} \\\Rightarrow C=\frac{\epsilon_0 A}{d}=\frac{\epsilon_0 \kappa A}{d} \\for~\kappa~is~relative~permeability~of~the~material~between~two~parallel~plates.$$

It seems that the capacitance of the capacitor depends on the dielectric and the distance between two parallel plates, and the material we use as the parallel plates are irrelevant as long as they conduct. Was my assumption correct?

Capacitance of a capacitor only depends on the geometry of the conductors and the relative permittivity of all space $\kappa (\mathbf{r})$.