# What is the capacitance of a superconductor?

I heard somewhere that the capacitance of a superconductor is much higher than regular conductors, but I haven't heard or seen anything yet proving or disproving such. So whats the truth? And why?

• You heard wrong. The capacitance is a geometrical thing. – Ron Maimon Dec 23 '11 at 4:50
• If that were true then how come electrolytic capacitors have more capacitance than regular non electrolytic capacitors? – Corbs Dec 23 '11 at 22:49
• It isn't an if, it just is true. You can increase the capacitance by using a dielectric in the space between the conducting objects holding the charge, but the capacitance has nothing to do with the material that is conducting the charge. – Ron Maimon Dec 24 '11 at 0:56
• Wait, then if that were true then if you raised the temperature of a capacitor then the capacitance should not change, but it does. – Corbs Dec 31 '11 at 3:51
• It's not an IF, it just is true. The temperature dependence is only of the dielectric constant of the dielectric material. There is no effect of the conducting material. – Ron Maimon Dec 31 '11 at 11:43

Capacitance is related to the total charge, $q$ built up on two metal plates and the voltage difference, $V$, between them (see http://en.wikipedia.org/wiki/Capacitance): $$C=\frac{q}{V}.$$ So, the capacitance should be independent of the inherent conductivity of the metal.
However, the time response will be different (see http://en.wikipedia.org/wiki/RC_circuit#Natural_response). There is intrinsic resistance in regular metals, which leads to a small, but non-zero charge/decay time constant $$\tau=RC$$ where $R$ is the intrinsic resistance and $C$ is the capacitance. If the metal is a superconductor, then $R$ is effectively $0$ and the capacitor charges and discharges instantly.
• The assumptions behind these equations is that the two plates can be connected by an infinitesimal perfectly conducting wire. However, in reality, no electrical signal can travel faster than the speed of light, so figure for a capacitor area of $A$ and plate separation of $d$ something like $\tau$ on the order of $\sqrt{A}d/c$. – Eli Lansey Jan 6 '12 at 11:51
• Every real component has some amount of self-inductance $L$. This gives the limit on how fast the current $I$ can start to flow. Let us assume you apply the voltage $U$ to the capacitor, then the current flow will increase linearly with $dI/dt = U/L$. – Martin J.H. Jun 26 '13 at 8:30