Why does capacitance not depend on the thickness of the capacitor plates?

So the formula for capacitance (being $C=\frac{A\,\epsilon_{0}\,\epsilon_{r}}{d}$) shows that the capacitance of a capacitor depends on the surface area of the capacitor plates.

As I understand it, this is because if the plates are larger, then for a given potential difference between the plates more electrons can be pushed onto the negative plate by the cell.

My question is, then by the same (and I am guessing flawed) logic, why does the thickness of the plates not affect the capacitance of the capacitor?

Or, put another way, why is the formula for capacitance not $C=\frac{v\,\epsilon_{0}\,\epsilon_{r}}{d}$, with $v$ being the volume of the capacitor plates?

Many thanks.

• The capacitor plates are conductors. All of the charge resides on the outer surface of a conductor. – David White Feb 10 '18 at 17:34
• So can there not be an excess of charge in the centre of a conducting material? If so, why? – Benjamin Rogers-Newsome Feb 10 '18 at 19:04
• the thickness does effect the capacitance; the 1st formula you are quoting is for an infinitely thin plate, in practice the formula is good if the plates' thickness is much less than their separation. The 2nd formula including the volume "$v$" cannot work, it is wrong even dimensionally. – hyportnex Feb 10 '18 at 19:06
• @hyportnex What is the relationship between the thickness and the capacitance of a capacitor? – Benjamin Rogers-Newsome Feb 10 '18 at 19:11
• There is an enormous literature of how plate thickness affects capacitance going back at least 120 years, but there is no simple formula for they all involve elliptic integrals and their approximations. Search for "microstrip capcitance", for example researchgate.net/publication/… – hyportnex Feb 11 '18 at 16:53