Shown above is a pair of circular aluminum plates connected to an electroscope (one to the case and one to the electrode). Rub one of the plastic rods with the cloth, and then stroke it on the electrode of the electroscope. With the plates as shown above, most of the charge you deposit stays on the electroscope, and the needle deflects. As you slide the moveable plate closer and closer to the fixed plate, the reading on the electroscope decreases, which indicates that charge has gone from the electroscope to the plates. If you slide the plate as close as possible to the fixed plate, the electroscope reading goes almost to zero. If you then increase the plate separation, the reading increases, indicating that charge has gone from the plates to the electroscope. This shows that the capacitance of the pair of plates is inversely proportional to the distance between them (vide infra).

The pair of plates connected to the electroscope in the photograph above constitute an arrangement or device called a capacitor. A capacitor can be any pair of insulated electrodes that carry equal and opposite charges, but the behavior of such an arrangement is much easier to describe when the electrodes are the same shape, and their shape and orientation result in a uniform electric field between them. Such an arrangement is a parallel plate capacitor, such as the one used in this demonstration. It consists of two aluminum discs whose diameter is 20.0 cm, and which can be set as far apart as about 12.8 cm, and brought as close together as 1 mm.

If we connect the plates to opposite sides of a battery or power supply, this results in the deposition of equal and opposite charges on the two plates. That is, one plate ends up with a charge of +q, and the other ends up with a charge of -q. The net charge on the capacitor remains zero, but each plate now has some charge, q, on it, which it did not have before we connected it to the battery or power supply, and this is the charge we use when we describe the behavior of the capacitor. As we might imagine, the magnitude of q depends on the potential difference we establish between the two plates (and which remains when we disconnect the power supply or battery), and is related to the potential difference by the equation

q = CV

where V is the potential difference between the plates, and C is a property of the capacitor called the capacitance. This is the quantity of charge that one can deposit on the plates of a capacitor by applying a particular voltage across it (or the potential difference that develps between the two plates for a given quantity of charge deposited on them). Its units are coulombs/volt, for which the single unit is called the farad (F); 1 farad = 1 coulomb/volt (= 1 C²/(N · m)). We are assuming that the plates are not surrounded by any material, that is, that they are in a vacuum. (That they are surrounded by air, however, makes only a very small difference.)

For flat rectangular or circular plates, if we ignore fringing effects near the edges, which are small if the plates are large compared to the distance between them, the electric field is uniform. It is also constant, and the electric flux, Φ_E equals EA, where E is the magnitude of the electric field, whose units are N/C, and A is the area of the plates. According to Gauss’s law,

ε₀Φ_E = q, and ε₀EA = q

where ε₀, the permittivity constant, equals 8.854 × 10^-12 C²/(N · m²). We can think of the work required to move a test charge q₀ from one plate to the other either in terms of the potential difference between the plates, in which case it equals q₀V, or in terms of the force exerted on it by the electric field over the distance between the two plates, which is q₀Ed, where d is the distance between the two plates. Since these must be equal, we have

V = Ed

If we substitute these into the first equation above, we get for the capacitance

C = q/V = (ε₀EA)/(Ed) = (ε₀A)/(d)

We see that the capacitance depends on both the area of the plates and the distance between them. This equation would look different for capacitors whose electrodes were not parallel plates, but in general, the greater the area of the electrodes and the closer they are to each other, the greater the capacitance. (See, for example, demonstration 60.17 -- Theremin.) This also explains the behavior of this demonstration. When you deposit charge on the top electrode of the electroscope, the charge distributes itself over the electrode and needle mechanism, and the capacitor plate that is connected to it, while an equal and opposite charge accumulates on the case of the electroscope and the on the opposite plate of the capacitor. The electroscope is itself a capacitor (of fixed capacitance), so the system comprises two capacitors in parallel. The potential difference across the electroscope and the potential difference between the two plates must be the same, which means that

q_electroscope = C_electroscopeV and q_plates = C_platesV, and

q = (q_electroscope + q_plates) = (C_electroscope + C_plates)V

We see that for a given charge deposited in the system,

V = q/(C_electroscope + C_plates)

With the plates at maximum separation as shown above, the total capacitance of the system is at its minimum. As you move the sliding plate closer and closer to the fixed plate, you increase the capacitance of the plates more and more. By the time the plates are at their minimum distance, their capacitance is large compared to what it was when they were at maximum separation. With the plates at maximum separation, then, if you deposit enough charge in the system to develop a voltage across it of several kilovolts (in the photograph, the scale reads just over 7 kilovolts), when you then slide the moveable plate closer to the fixed plate, the increasing capacitance of the plates causes a decrease in the voltage across the system. As you continue decreasing the distance between the plates, their capacitance becomes greater and greater, and the voltage continues to decrease. When the plates are at their minimum distance, their capacitance is large compared to that of the electroscope, the total capacitance of the system is much larger than before, and the voltage across the system is much lower than before. Now, if you slide the moveable plate away from the fixed plate, the capacitance of the plates begins to decrease, the voltage across the system begins to increase, and the needle of the electroscope begins to swing upward. When the plates are again at their maximum separation, the electroscope reads the same voltage it had when you first charged the system, or at least close to it, depending on whether any significant quantity of charge has leaked out of the system. At all times, the portion of the total charge that resides in each part of the system is in direct proportion to its capacitance. For example, if the capacitance of the plates equals that of the electroscope, they hold equal quantities of charge. If the capacitance of the plates is twice that of the electroscope, the plates hold two-thirds of the charge, and one-third of the charge is on the electroscope.

For reference, the capacitance of the electroscope is about 29.1 pF, the capacitance of the plates at maximum separation is about 20.3 pF, and the capacitance of the plates when gently closed is about 230 pF. So the minimum capacitance is about 49.4 pF, when the plates have 41 percent of the total capacitance, and the maximum capacitance is about 259 pF, when the plates have 89 percent of the total capacitance.

References:

1) Halliday, David and Resnick, Robert. Physics, Part Two, Third Edition (New York: John Wiley and Sons, 1978), pp. 650-655.
2) Sears, Francis Weston and Zemansky, Mark W. College Physics, Third Edition (Reading, Massachusetts: Addison-Wesley Publishing Company, Inc., 1960), pp.515-521.