Capacitance and conductors

Question

The capacitance of a sphere of radius a is : $C=Q/V=1/4\pi\epsilon_0a$. If the charge density is assumed on its surface $Q=\rho_s4\pi a^2$, this gives a charge density $\rho_s$:

$\rho_s = \frac{\epsilon_0}{a} V ~~(C/m^2)$

The smaller the sphere, the smaller the charge density to obtain a uniformm potential V, hence the smaller the capacitance.

However, this analysis is valid regardless of the nature of the material which could also be an insulator. Is the capacitance a purely geometrical feature?

Where does the nature of the material comes into account? Is it related to the amount of work necessary to drag the electrons from the material to its surface? Where conductors would require less work than insulator?

Answer 1

Your formula for the capacitance of the sphere, i.e., the relation between the voltage $V$ and the charge $Q$ on its surface, is correct and identical to the case of a conductive sphere because you assume that the charge is evenly distributed with density $\rho_s$ on the surface. Due to Gauss law, the formula would also be correct if the charge were a point charge in the center of the sphere or distributed in the sphere with spherical symmetry. This is completely independent of the material. In general, however, you need a conductor for a capacitor. If the sphere were an insulator, the charge would not (re-) distribute evenly over (the surface of) the sphere when charge is transported to or from a point on the surface. Thus the capacitance formula would not be valid any longer and you could not use the sphere as a capacitor because you could not charge and discharge it into a circuit easily.

Answer 2

Indeed capacitance is a purely geometric feature! You are spot on in your statement that indeed capacitance is related to the work required to drag an electron from one surface to another. This is seen from the definition of potential as work per unit charge. That is, $$C=\frac{Q}{V}=\frac{Q}{W/q_0} \Rightarrow C\propto\frac{1}{W}.$$ Hence we can think of the capacitance (per unit test charge) as being an inverse measure of how much work has to be done to assemble the charges. Now since $W$ itself is always proportional to the total charge $Q$, you can count that it will always cancel, leaving nothing but factors that depend on the geometry of the configuration.

As for the question about the material; people would never use insulators themselves as capacitors in the sense that I believe you are suggesting. Rather the surfaces (or "plates") that bound our region where the electric field is, are always made of conducting material. What usually is the case is that we insert an insulating material between the two surfaces. What this does is reduce the electric field between the plates and thereby increases the capacitance. Mathematically, this amounts to (for linear materials) taking $$\epsilon_0\rightarrow\epsilon:= \epsilon_0 \epsilon_r$$ in all of your equations. Where $\epsilon_r$ is called the "relative permeability" or "dielectric constant" (sometimes you also see this called $\kappa$. So this is how we account for the capacitance where we have different materials sandwiched between the plates.