Consider a system of two insulated conductors placed in such a way that if charged on one conducted is +Q, the induced charged on the Other conductor will be -Q. If V is the potential difference between the conductors, for a particular system, the ratio of Q to V is constant, depending on the size and shape of electric field and the medium in which the force due to field exists. This proportionality constant is called capacitance. “The capacitance of a system is the ratio of its charge to its potential.”
Here Q is expressed in coulombs, V is expressed in volts and the capacitance C is given in farads.
Farad being a very large unit, in practice the smaller unit microfarad is used to express the capacitance of the system.
1 microfarad() = 10-6 farad () &
1 picofarad = 10-12 farad
Parallel plate capacitor: the following figure shows two parallel metallic plates separated by a distance d. When the potential difference V is applied across these metallic plates, a positive charge + Q1 is formed on the upper plate and the negative charge – Q1 is formed on the lower plate.
Now if the sheet of Bakelite or the slab glass or any other suitable dielectric is inserted between the plates to fill the gap between them, the charges on the plate will increase to +Q2 and -Q2 with the same potential difference V.
The increase in charge is due to the presence of the dielectric medium between the plates. The ratio Q2 by Q1 is termed as the relative permittivity or dielectric constant of the dielectric between the plates.
The relative permitivities of some of the dielectrics commonly used are given in table
| Medium | Relative permittivity |
| Air | 1.00059 |
| Glass | 5 to 10 |
| Bakelite | 4.5 to 5.5 |
| Ebonite | 2.8 |
| Mica | 2.0 to 6.5 |
| Paper | 2.0 to 2.6 |
| Paraffin | 2.1 to 2.5 |
| Rubber | 2.0 to 3.5 |
| Oil | 2.2 to 4.8 |
| Wood | 2.5 to 7.5 |
Example:- A capacitor consist of two parallel plates in air. Each plate has an area of 1 sq. meter and the distance between the plates is 0.5 cm. If the capacitance is 1.5 * 10-3 micro farads, what quantity of electricity is displaced when potential between the plates is 15,000 volts?
Solution:- C = Q/V or Q = C*V
Q = 1.5 *10-3 * 10-6*15000
= 2.25 * 10-5 coulomb
In a parallel plate capacitor, if it is assumed that the field strength is uniform over the area of the plates, the field strength can be represented by σ/εε0
where σ is the surface density of the charge.
If A is the plate area,
σ = Q/A & E = Q/εε0 A
If d is the distance between the plates, the potential difference V between the plates is given by V = Ed
Substituting in relation…
V = Ed = Qd/ εε0A
and Capacitance C = Q/V = εε0A/d farads
From this relation, it it easy to find out the area of plates of capacitor required to obtain a certain value of capacitance.
A = Cd/εε0