The charge, Q, on the plates and the voltage, V, between the plates are related according to the equation where C is the capacitance which depends upon the geometry and dimensions of the capacitor. For a parallel plate capacitor with plate area A and separation d, its capacitance is ε A
The capacitance of the electrometer and cable is approximately 20 pF (1 pF = 1 pico Farad = 10-12 Farad). The experimental result should be close to this value. (4 pts) Why is it sufficient to add charges to only one plate?
However, it is a misnomer to think that the capacitance of a capacitor is defined by the amount of charge and voltage. Capacitance is defined by the geometry of the capacitor design, or particularly on the cross sectional area of the plates and the separation distance of the plates (and also the material, if any, placed between the plates).
The capacitance C1 of the electrometer and cable is in parallel with the test capacitor of capacitance C2. Therefore, C1 can be calculated according to The capacitance of the electrometer and cable is approximately 20 pF (1 pF = 1 pico Farad = 10-12 Farad). The experimental result should be close to this value.
When a voltage V is applied to the capacitor, it stores a charge Q, as shown. We can see how its capacitance may depend on A and d by considering characteristics of the Coulomb force. We know that force between the charges increases with charge values and decreases with the distance between them.
A simple capacitor is the parallel plate capacitor, represented in Figure 1. The plates have an area A and are separated by a distance d with a dielectric ( ) in between. The plates carry charges +Q and Q, respectively, on their surfaces. The capacitance of the parallel plate capacitor is given by