Chapter 5: Electric Fields in Material Space

Example 5.1, Page number: 167

In :

import scipy
import scipy.integrate

#Variable Declaration

r1= 0.2                # radius of hemispherical shell in metres
r2= 0.1                # radius of spherical shell in metres

#Calculations

#Calculation of current through hemispherical shell

def J1(phi,theta):
s1=(1/r1)*(2* scipy.cos(theta)* scipy.sin(theta))
return s1

if __name__ == '__main__':

I1, error = scipy.integrate.dblquad(lambda theta , phi: J1(phi,theta),
0, 2*scipy.pi, lambda theta: 0, lambda theta: scipy.pi/2)

#Calculation of current through spherical shell

def J2(phi,theta):
s2=(1/r2)*(2* scipy.cos(theta)* scipy.sin(theta))
return s2

if __name__ == '__main__':

I2, error = scipy.integrate.dblquad(lambda theta , phi: J1(phi,theta),
0, 2*scipy.pi, lambda theta: 0, lambda theta: scipy.pi)

#Results

print 'Current through hemispherical shell=',round(I1,1),'A'
print 'Current through spherical shell=',round(I2,0),'A'

Current through hemispherical shell= 31.4 A
Current through spherical shell= 0.0 A

Example 5.2, Page number: 168

In :

#Variable Declaration

ps=10**-7     #Surface charge density of the belt in Couloumb/metre^2
u=2           #Speed of the belt in metres/sec
w=0.1         #Width of the belt in metres
t=5           #Time taken in seconds

#Calculations

I=ps*u*w      #Current in amperes
Q=I*t*10**9   #Charge collected in 5 seconds in nano Coloumbs

#Result

print "The charge collected in 5 seconds is ",Q,"nC"
The charge collected in 5 seconds is  100.0 nC

Example 5.3, Page number: 169

In :

#Variable Declaration

n=10**29              #Number density of electrons in m^-3
e=-1.6*10**-19        #Electronic charge in Coloumbs
sigma=5*10**7         #Current density in S/m
E=10**-2              #Electric Field in V/m
S=(3.14*10**-6)/4     #Cross sectional area of the wire in m^2

#Calculations

pv=n*e                #Charge density of free electrons in C/m^3
J=sigma*E*10**-3      #Current density in kA/m^2
I=J*S*10**3           #Current in amperes
u=J*10**3/pv          #Drift velocity in m/s

#Results

print "The charge density is ",pv,"C/m^3"
print "The current density is ",J,"kA/m^2"
print "The current is ",round(I,3), "A"
print "The drift velocity is ",-u,"m/s"
The charge density is  -16000000000.0 C/m^3
The current density is  500.0 kA/m^2
The current is  0.393 A
The drift velocity is  3.125e-05 m/s

Example 5.4, Page number: 170

In :

import scipy

#Variable Declaration

l=4                            #Length of the lead bar in m
d=3                            #Width of the lead bar in cm
r=0.5                          #Radius of the hole drilled in cm
sigma=5*10**6                  #Conductivity of the bar in S/m

#Calculation

S=(d**2-(scipy.pi*r**2))       #Cross sectional area in cm^2
R=l/(S*sigma*10**-4)           #Resistance in ohms

#Result

print 'The resistance between the square ends is',round(R*10**6),'micro ohms'
The resistance between the square ends is 974.0 micro ohms

Example 5.6, Page number: 177

In :
import scipy

#Variable Declaration

e0=10**-9/(36*scipy.pi)         #permittivity of free space in Farad/m
er=2.55                         #relative permittivity (dimensionless)
E=10*10**3                      #Electric field in V/m
chi=er-1.0                      #Electric susceptibility (dimensionless)
d=1.5                           #Distance between plates in mm

#Calculations

D=e0*er*E*10**9                 #D in nC/m^2

P=chi*e0*E*10**9                #P in nC/m^2

ps=D                            #The surface charge density of
#free charge in nC/m^2

pps =P                          #The surface charge density of
#polarization charge in nC/m^2

V=E*d*10**-3                    #The potential difference between
#the plates in volts

#Results

print 'D =',round(D,2),'nC/m^2'
print 'P =',round(P,0),'nC/m^2'
print 'Surface charge density of free charge =',round(ps,2),'nC/m^2'
print 'Surface charge density of polarization charge =',round(pps,0),'nC/m^2'
print 'The potential difference between the plates =',V,'V'
D = 225.47 nC/m^2
P = 137.0 nC/m^2
Surface charge density of free charge = 225.47 nC/m^2
Surface charge density of polarization charge = 137.0 nC/m^2
The potential difference between the plates = 15.0 V

Example 5.7, Page number: 178

In :

import scipy

#Variable Declaration

e0=10**-9/(36*scipy.pi)                   #permittivity of free space

er=5.7                                    #relative permittivity
#(dimensionless)

chi=er-1                                  #Electric susceptibility
#(dimensionless)

r=0.1                                     #radius of sphere in m

q1=2                                      #charge on sphere in pC

q2=-4                                     #value of point charge in pC

#Calculations

E=q1/(4*scipy.pi*e0*er*r**2)              #Electric field on the
#sphere in pV/m

P=chi*e0*E                                #Polarisation in pC/m^2

pps=P                                     #The surface density of polarization
#charge in pC/m^2

F=(q1*q2*10**-12)/(4*scipy.pi*e0*er*r**2) #Force exerted on point charge in pN

#Results

print 'The surface density of polarization'
print 'charge on the surface of the sphere =',round(pps,2),'pC/m^2'
print 'Force exerted on -4 pC charge =',round(F,3),'pN in the radial direction'
The surface density of polarization
charge on the surface of the sphere = 13.12 pC/m^2
Force exerted on -4 pC charge = -1.263 pN in the radial direction

Example 5.9, Page number: 188

In :

#Variable Declarartion

import scipy
from numpy import *

an=array([0,0,1])            #Unit vector normal to the interface
E1=array([5,-2,3])           #Electric field for z >=0 in kV/m
e_r1=4                       #Relative permittivity for z >=0 (dimensionless)
e_r2=3                       #Relative permittivity for z <=0 (dimensionless)
e0=(10**-9)/(36*scipy.pi)    #Permittivity of free space in Farad/m
V=2*2*2                      #Volume of cube placed in region 2 in m^3

#Calculations

E1n=array([0,0,dot(E1,an)])   #The normal component of E1 in kV/m
E1t=E1-E1n                    #Transverse component of E1 in kV/m
E2t=E1t                       #Transverse component of E2 in kV/m
E2n=e_r1*E1n/e_r2             #Normal Component of E2 in kV/m
E2=E2n+E2t                    #The total field E2 in kV/m

theta1= 90- 180*scipy.arccos(dot(E1,an)/        #Angle between E1 and
scipy.sqrt(dot(E1,E1)))/scipy.pi   #interface in degrees

theta2= 90- 180*scipy.arccos(dot(E2,an)/        #Angle between E2 and
scipy.sqrt(dot(E2,E2)))/scipy.pi   #interface in degrees

We1= 0.5*e0*e_r1*dot(E1,E1)*10**6       # The energy density of E1 in J/m^3
We2= 0.5*e0*e_r2*dot(E2,E2)*10**6       # The energy density of E2 in J/m^3
W= We2*V                                # The energy within the cube in J

#Results

print 'The electric field for the region z <=0 is',E2,'kV/m'
print 'The angle E1 makes with the boundary is',round(theta1,1),'degrees'
print 'The angle E2 makes with the boundary is',round(theta2,1),'degrees'
print 'The energy density in dielectric 1 is',round(We1*10**6,0),'J/m^3'
print 'The energy density in dielectric 2 is',round(We2*10**6,0),'J/m^3'
print 'The energy within the cube is',round(W*1000,3),'mJ'
The electric field for the region z <=0 is [ 5 -2  4] kV/m
The angle E1 makes with the boundary is 29.1 degrees
The angle E2 makes with the boundary is 36.6 degrees
The energy density in dielectric 1 is 672.0 J/m^3
The energy density in dielectric 2 is 597.0 J/m^3
The energy within the cube is 4.775 mJ

Example 5.10, Page number: 190

In :

import scipy

#Variable Declaration

e=(10**-9)/(36*scipy.pi)     #Permittivity of free space in Farad/m
er=2                         #Relative permittivity (dimensionless)
ps=2                         #Surface charge in nC/m^2

#Calculations

#Point A is in the region y <=0. Hence E=D=0
#For point B which is in the region y >=0,

Dn=ps                        #Displacement current in nC/m^2
En=Dn*10**-9/(e*er)          #Electric Field

#Result

print 'E at point A= 0'
print 'D at point A= 0'
print 'E at point B=',round(En,2),'V/m along positive y direction'
print 'D at point B=',Dn,'nC/m^2 along positive y direction'
E at point A= 0
D at point A= 0
E at point B= 113.1 V/m along positive y direction
D at point B= 2 nC/m^2 along positive y direction