Capacitance is the efficiency of storing a charge difference between two conductors at the expense of the voltage between them. A high capacitance means there is a lot of charge stored per volt between the conductors. This is described by the definition, C = Q/V. If the definition requires two conductors, what does it mean to talk about the capacitance of one conductor?
If you put some excess charge on a conductor, sitting in space, it will rise to a new voltage, compared to any other place measured as the reference. It will have a capacitance, the ratio of the charge added to its change in voltage.
It’s difficult to calculate the voltage generated on an isolated conductor except for simple geometries, like a sphere. The case of two concentric spheres is a classical problem in all freshmen EM classes. A great explanation is found here.
Suppose we make the outer sphere bigger and bigger, effectively moving it farther and farther away. The value of b gets larger and larger and 1/b goes to zero. If the outer sphere is more than 10 x the radius of the inner sphere, the value of 1/b is less than 10% the value of of 1/a and has only a small impact. In this extreme case, when the outer sphere is very far away- like to the floor, or the walls of the room, the capacitance of the inner sphere, to any metal far away, is related to the size of the smaller sphere, a.
The capacitance of a small, isolated sphere is just 4 x pi x epsilon zero x a. Using values of 0.225 pF/inch for epsilon zero, the capacitance of a sphere, with radius a is: C in pF = 1.4 x D, with D the diameter of the sphere in inches.
This is a startling result. It says that a piece of metal floating in space has a capacitance to any far away surface and it is roughly related to its diameter. If it is other than a sphere, it is a little hard to calculate, so to use the luxury of a simple estimate, we have to assume it is a spherical shape.
For example, if we have a cable sticking out from a computer that is 3 inches long, maybe its equivalent to a sphere with a diameter of about 1 inch. By nature of its size, it will have a capacitance to any other surface, far away, of about 2 pF.
This is a very good estimate of the capacitance between the shield of a cable to the floor. It’s the fringe electric fields of external cables to the floor that is the return path for common currents on the cable. It’s the impedance of this path that usually determines how much common current flows on the cable shield.
If there is 2 pF of fringe field capacitance, at 100 MHz, the impedance is roughly 1 kOhm. If the ground bounce noise on a plane, that drives the common currents, is just 100 mV, the common currents will be on the order of 0.1 v/1k Ohm = 100 uA. It only takes 3 uA of common current to fail an FCC class B certification test so we see how easy it is for ground bounce to cause EMC problems.