In the photograph above is a Leybold parallel plate capacitor connected to a Global Specialties® capacitance meter. The plates of the capacitor are 25.8 cm (10.2″) in diameter, and you can vary their separation by means of the knob at left in the photograph. The meter shows the capacitance, which changes as you vary the plate separation. You can insert one or both of the two sheets of plastic (sitting in front of the capacitor in the photograph) to show that a dielectric inserted between the plates increases the capacitance. A video camera connected to the data projector allows you to show the meter display to the class.
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. As noted above, this is a Leybold parallel plate capacitor, which has two circular plates that are 25.8 cm (10.2″) in diameter. One plate is moveable, and is electrically connected to the frame of the apparatus. The fixed plate is mounted to the frame via an insulator. The knob at left in the photograph operates a screw drive, by which you can adjust the distance between the plates anywhere from about 2 mm to about 70 mm. (An adjustable stop allows you to set the minimum separation smaller than this, but 2 mm is the current setting.) A scale on top of the housing of the screw drive (visible in the photograph) gives a direct reading of the plate separation.
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, ΦE 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,
ε0ΦE = q, and ε0EA = q
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.) From this equation, we can get an idea of the size of the unit of capacitance by calculating for a pair of plates separated by a certain distance, the area necessary for them to have a capacitance of one farad. If we set the plates 1 mm apart, we get
A = Cd/ε0 = (1.0 F)(1.0 × 10-3 m)/8.854 × 10-12 (C2/(N · m2)) = 1.1 × 108 m2
If the plates are square, this corresponds to a pair of plates whose sides are 10,600 m, or a little over 6.5 miles long. The farad is a very large unit. As a result, depending on the size of the capacitor, the values of most capacitors are usually given in microfarads (μF or uF), nanofarads (nF) or picofarads (pF).
The plates of the Leybold capacitor have an area of 5.23 × 10-2 m2, so ε0A = 4.63 × 10-13 C2/N (= F · m). For separations from 70 mm to 2 mm, then, we should expect that the capacitance of this apparatus ranges from tens to hundreds of picofarads. At a separation of 20 mm the meter reads a capacitance of about 72 pF (0.072 nF). As you move the moveable plate closer to the fixed plate the capacitance increases. When the separation is 2 mm, the meter shows a capacitance of 272 pF (0.272 nF). (The capacitances you would calculate from the equation above for these separations are 23 pF and 231 pF, respectively.)
As noted above, we have considered the capacitor with no material between the plates. It turns out that if you put an insulating material, or dielectric, between the plates, this increases the capacitance. When you charge the capacitor with a dielectric between the plates, the dielectric is polarized in the electric field between the plates. Though the dielectric does not conduct electricity, the charges within it can redistribute themselves in the direction of the electric field. Negative charges concentrate near the positive plate of the capacitor, and positive charges concentrate near the negative plate. As a result, an electric field is set up across the dielectric, which opposes the electric field produced by the charges on the capacitor plates. If we call the electric field produced by the plates E0, and the electric field across the dielectric E′, then the resultant field, E, equals E0 + E′. Since E′opposes E0, E is smaller than E0, but still points in the same direction. Since V = Ed, for a given charge q in the capacitor, the voltage across the capacitor is lower with the dielectric inserted than without it, in proportion to E:
E0/E = V0/V = κ
where κ is the dielectric constant of the material. (If we held the voltage across the capacitor constant at V, then the charge would be greater with the dielectric than without it, in proportion to the dielectric constant. That is, q/q0 = κ. Also, this equation holds only when the dielectric fills the gap between the plates over their entire area.) For capacitor plates in a vacuum, κ = 1. For air, κ = 1.00054, so in our analysis of the capacitor without a dielectric, unless we are concerned with errors of less than one part per thousand, we need not worry that our capacitor has an air dielectric.
As noted above, two sheets of dielectric material are available for use with the capacitor. These are plastic sheets that are 22.3 cm square. One sheet is 0.8 mm thick, and the other is 1.7 mm thick. For maximum effect, they must fill the gap between the plates, and should cover the entire area of the plates. If they don’t fill the gap between the plates, the field across them is smaller in proportion to that across the plates than it would be if they filled the gap. Though these squares are smaller than the diameter of the plates, they cover most of the area, so their effect is still large. With the plates set 2 mm apart, sliding the thinner sheet between them raises the capacitance to about 385 pF, and inserting the thicker sheet raises it to about 680 pF. With the plates far enough apart to slide both sheets in, the capacitance is 230 pF with no sheets inserted, 285 pF with the thinner sheet inserted, 394 pF with the thicker sheet inserted, and 700 pF with both sheets inserted. If you squeeze the sheets by screwing the moveable plate toward the fixed plate, you can increase the capacitance to 950 pF.
References:
1) Halliday, David and Resnick, Robert. Physics, Part Two, Third Edition (New York: John Wiley and Sons, 1978), pp. 650-662.
2) Sears, Francis Weston and Zemansky, Mark W. College Physics, Third Edition (Reading, Massachusetts: Addison-Wesley Publishing Company, Inc., 1960), pp.515-521.