Chapter 6: Electrostatic Boundary Value Problems

Example 6.12, Page number: 238

 
import scipy

#Variable Declaration

e0=10**-9/(36*scipy.pi)    #Permittivity of free space in Farad/m
er1=4                      #Relative permittivity of material 1 (dimensionless)
er2=6                      #Relative permittivity of material 2 (dimensionless)
d=5*10** -3                #The spacing between the capacitor plates in m
S=30*10** -4               #The surface area of the plates

#Calculations

#Calculation for Capacitor 1

C1=e0*er1*S*2/d            #Capacitance in Farads
C2=e0*er2*S*2/d            #Capacitance in Farads
Ca=C1*C2/(C1+C2)           #Total capacitance in Farads

#Calculation for Capacitor 2

C1= e0*er1*S/(2*d)         #Capacitance in Farads
C2= e0*er2*S/(2*d)         #Capacitance in Farads
Cb= C1+C2                  #Total capacitance in Farads


#Results

print 'The capacitance of capacitor 1 =',round(Ca*10**12,2),'pF'
print 'The capacitance of capacitor 2 =',round(Cb*10**12,2),'pF'

The capacitance of capacitor 1 = 25.46 pF
The capacitance of capacitor 2 = 26.53 pF