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 Quote by Simon Bridge Please show the derivation for the capacitance of a spherical conductor by itself. Perhaps you are thinking of self capacitance ?
You can google "capacitance of an isolated sphere" and "charge on an isolated sphere" for all kinds of examples, but here one I've lifted from the bunch because it actually states, "Note there is only one surface, but the formula still works" which emphasizes that there's no second plate. This example goes like this:

 Quote by example web hit Isolated Charge A metal sphere with a 0.25 m radius is in a vacuum. Find the capacitance. Note there is only one surface, but the formula still works. The potential of the sphere is V=kQ/r=Q/(4 pi epsilon_naught r) The capacitance is C=Q/V=4 pi epsilon_naught r=r/k For the sphere C = (0.25 m) / (8.99 x 10^9 Nm^2/C^2) C = 28 x 10^-12 F = 28 pF.
If you have been told the voltage on the generator, you can compute the charge using the above formulas, however as you can see, if the voltage and charge are unknown, they cancel out when you substitute the equation for V into that for C. In other words, the value C is independent of the current charge and potential on the sphere. If you have been given the voltage (as you will find in many example hits) then you can calculate the charge as Q=CV.

When I was first introduced to this idea of the capacitance of an isolated body, it was decades after I had been working as an electronic designer (and I have two patents) where I always thought of capacitors as the "two plate" variety, and so I had considerable trouble initially understanding the concept of the capacitance of an isolated body in space. However it does make sense when you think of the following example: Suppose you have a signal generator with its ground lead connected to ground and its hot lead connected to an isolated coin such as a dime. If you connect an oscilloscope up from ground to the coin clearly you will read the signal. This shows that the coin must have a potential that "follows" the signal generator wrt ground. However the coin is small, then the signal generator doesn't have to supply much current to "charge" and "discharge" the coin. But what if you connect the signal generator up to a huge metal object like a metal ship that's in dry dock and isolated from ground. If you connect the signal generator up to one side of the ship and run over to the opposite side with your oscilloscope, the only way you still still see the test signal is if the signal generator can supply sufficient current to "charge up" and "charge down, i.e. discharge" the body of the entire ship. Intuitively, you can see that the ship is huge and a lot of electrons need to be pushed onto and pulled off of the surface in order that the scope will actually show the test voltage. Therefore, the large ship represents a "capacitive load" to the signal generator, and in fact its a capacitive load whose value could be approximated as a huge metal sphere. If the ship is too large, it will have too much capacitance, and the signal generator will not be able to provide sufficient current to charge and discharge the ship, and the scope will not show the full amplitude of the signal. In practice, as an electronic designer I would never have considered such issues, because my circuits use small short wires to connect points to other points and the capacitance of those isolated wires are so negligible in most applications (except in high frequency microwave circuits) that we simply ignore this effect. But you can certainly see that its a completely different ball game if one of your circuit wires is the size of a ship! You can no longer just view it as a conductive path with zero impedance in that case.

Of course you don't need such an elaborate set up to test this. It will do to connect a high resistor with a high frequency sine wave. Monitor the other end of the resistor with a scope, and then have someone electrically isolated from all the equipment touch that end of the resistor. The amplitude of the signal on the scope will drop significantly. The human body capacitance to a far ground is at least 100-200 pF. If you use a 100K resistor and set the signal generator for 10KHz. If you set the amplitude to 1vpp, then with the human touching the wire (and remember they are otherwise isolated from ground and from the circuit), the amplitude of the signal as measured by the scope will drop to about 60% of its original value. If you repeat the experiment with a small child, it will only drop to about 80%. And if you repeat it using a small coin, the amplitude will not change noticeably. You can try different objects of different sizes to see the effect. In all cases, the objects act as a capacitor to ground and have the exact same effect as though you really connected up a two plate capacitor of that value between the end of the resistor and ground. I know its weird, but I've done the experiment, so I know its true.