where C is a positive proportionality constant called capacitance.Physically capacitance is a measure of the capacity of storing electric charge for a given potential difference AV.The SI unit of capacitance is the farad (F): 1 F=1 farad=1 coulomb/volt=1 C/V A typical capacitance is in the picofarad (1 pF=10-2F)to millifarad range, (1mF=10-3F=1000F;1F=10-6F) Figure 5.1.3(a)shows the symbol which is used to represent capacitors in circuits.For a polarized fixed capacitor which has a definite polarity,Figure 5.1.3(b)is sometimes used. (a) (b) Figure 5.1.3 Capacitor symbols. 5.2 Calculation of Capacitance Let's see how capacitance can be computed in systems with simple geometry. Example 5.1:Parallel-Plate Capacitor Consider two metallic plates of equal area 4 separated by a distance d,as shown in Figure 5.2.1 below.The top plate carries a charge while the bottom plate carries a charge-O.The charging of the plates can be accomplished by means of a battery which produces a potential difference.Find the capacitance of the system. Figure 5.2.1 The electric field between the plates of a parallel-plate capacitor Solution: To find the capacitance C,we first need to know the electric field between the plates.A real capacitor is finite in size.Thus,the electric field lines at the edge of the plates are not straight lines,and the field is not contained entirely between the plates.This is known as 3where C is a positive proportionality constant called capacitance. Physically, capacitance is a measure of the capacity of storing electric charge for a given potential difference ∆V . The SI unit of capacitance is the farad (F) : 1 F = = 1 farad 1 coulomb volt = 1 C V A typical capacitance is in the picofarad ( ) to millifarad range, ( ). 12 1 pF 10 F − = 3 6 1 mF 10 F=1000µ µ F; 1 F 10 F − − = = Figure 5.1.3(a) shows the symbol which is used to represent capacitors in circuits. For a polarized fixed capacitor which has a definite polarity, Figure 5.1.3(b) is sometimes used. (a) (b) Figure 5.1.3 Capacitor symbols. 5.2 Calculation of Capacitance Let’s see how capacitance can be computed in systems with simple geometry. Example 5.1: Parallel-Plate Capacitor Consider two metallic plates of equal area A separated by a distance d, as shown in Figure 5.2.1 below. The top plate carries a charge +Q while the bottom plate carries a charge –Q. The charging of the plates can be accomplished by means of a battery which produces a potential difference. Find the capacitance of the system. Figure 5.2.1 The electric field between the plates of a parallel-plate capacitor Solution: To find the capacitance C, we first need to know the electric field between the plates. A real capacitor is finite in size. Thus, the electric field lines at the edge of the plates are not straight lines, and the field is not contained entirely between the plates. This is known as 3