Chapter 14: Dielectric Properties of Materials

Exa 14.1

In [1]:
from __future__ import division
import math
 # Python Code Ex14.1 Polarization of water molecule: Page-456 (2010)
 
 
 
  
#Variable declaration


NA = 6.023e+23;                         # Avogadro's number
p = 6e-030;                         # Dipole moment of water molecule, C-m
r = 1e-03;                              # Radius of water molecule, m
M = 18e-03;                        # Molecular weight of water, kg
d = 1e+03;               # Density of water, kg per metre cube


#Calculation

V = M/d;                # Volume of water, metre cube
 # Now M/d metre cube volume will contain NA = 6.023e+023 water molecules, 
 #so that 4*%math.pi/3*(r**3) metre cube volume will contain 
N = NA*d*4*math.pi*r**3/(M*3);    # Number of water molecules per metre cube
P = N*p;     # Polarization of water molecules, coulomb per metre square



#Results

print"\nThe polarization of water molecules ="
print round((P*10**10),2)*10**-10,"coulomb per metre square"
 
The polarization of water molecules =
8.41e-10 coulomb per metre square

Exa 14.2

In [2]:
from __future__ import division
import math
 # Python Code  Ex14.2  Calculating  dielectric  constant  from  
 # electric  polarizability  of  the  atom:  Page-464  (2010)
 
 
  
#Variable declaration


#  Electric  polarizability  of  the  Kr-atom,  farad-metre  square
alpha_Kr  =  2.18e-040;
NA  =  6.023e+023;                                #  Avogadro's  number
#  Electrical  permittivity  of  free  space,
#  coulomb  square  per  newton  per  metre  square
epsilon_0  =  8.85e-012;        



#Calculation

N  =  NA/(22.4e-03);#  Number  of  Kr  atoms  per  metre  cube
#  Relative  electrical  permittivity  of  Kr  specimen
epsilon_r  =  N*alpha_Kr/epsilon_0  +  1;  


#Result
    
print"\nThe  diectric  constant  of  Kr  specimen  =",  round(epsilon_r,5)
The  diectric  constant  of  Kr  specimen  = 1.00066

Exa 14.3

In [3]:
from __future__ import division
import math
 # Python Code Ex14.3 Calculating electric polarizability of 
 #a molecule from its susceptibility: Page-464 (2010)
 
 
  
#Variable declaration


NA = 6.023e+023;                       # Avogadro's number
# Electrical permittivity of free space, coulomb square per newton per metre
epsilon_0 = 8.85e-012;                  
chi = 0.985e-03;       # Electrical susceptibility of carbon-dioxide molecule
rho = 1.977;              # Density of carbon-dioxide, kg per metre cube
M = 44e-03;                        # Molecular weight of CO2, kg



#Calculation

N = NA*rho/M;         # Number of molecules per unit volume, per metre cube
# Total electric polarizability of carbon-dioxide, farad-metre square
alpha = epsilon_0*chi/N;                


#Result

print"\nThe total electric polarizability of carbon-dioxide ="
print round((alpha*10**40),2)*10**-40,"farad-metre square"
 
The total electric polarizability of carbon-dioxide =
3.22e-40 farad-metre square

Exa 14.4

In [4]:
from __future__ import division
import math
 # Python Code Ex14.4 Calculating electric polarizability 
 #of Oxygen atom: Page-465 (2010)
 
 
  
#Variable declaration


e = 1.602e-019; # Charge on an electron, coulomb
p = 0.5e-022; # Dipole moment of oxygen atom, C-m
# Distnace of the centre of negative charge cloud from the nucleus, m
d = 4e-017; 
# Electrical permittivity of free space, coulomb square per newton per metre
epsilon_0 = 8.85e-012; 



#Calculation

 # In equilibrium, Coulomb interaction = Lorentz force
 # i.e. 8*e*E = (8*e)*(8*e)/(4*%math.pi*epsilon_0*d**2)
 # Solving for E
# The strength of local electric field, volt per metre
E = 8*e/(4*math.pi*epsilon_0*d**2); 
 # As p = alpha*E, solving for alpha
alpha = p/E; # Atomic polarizability of oxygen, farad-metre square




#Result

print"\nThe atomic polarizability of oxygen = "
print round((alpha*10**48),2)*10**-48,"farad-metre square"
The atomic polarizability of oxygen = 
6.94e-48 farad-metre square

Exa 14.5

In [5]:
from __future__ import division
import math
 # Python Code Ex14.5 Dipolar polarization of HCl molecule: Page-470 (2010)
 
 
 
  
#Variable declaration


k = 1.38e-023; # Boltzmann constant, J/mol/K
T = 300; # Temperature of the HCl vapour, kelvin
N = 1e+027; # Number of HCL molecuels per unit volume, per metre cube
# Electric field strength to which the HCL vapour is subjected, volt/m
E = 1e+06; 
p = 3.46e-030; # The dipole moment of HCl molecule,C-m



#Calculation

# Dipolar polarizability of HCl molecule, farad-metre square
alpha_d = p**2/(3*k*T);
 # As P = N*p = N*alpha_d*E
#Orientational or Dipolar polarization of HCl molecule,coulomb per metre square
P = N*alpha_d*E;
E_M = p*E; # Magnetic energy stored in the dipole-field system, joule
E_Th = k*T; # Thermal energy of the HCl molecule, joule
a = E_M/E_Th; # Ratio of magnetic energy to the thermal energy


#Result

print"\nThe orientational polarization of molecules in HCl vapour ="
print round((P*10**7),2)*10**-7,"coulomb per metre square"
print"\nThe ratio of magnetic energy to the thermal energy ="
print round(a,6)," << 1"
 
The orientational polarization of molecules in HCl vapour =
9.64e-07 coulomb per metre square

The ratio of magnetic energy to the thermal energy =
0.000836  << 1

Exa 14.6

In [6]:
from __future__ import division
import math
#Python Code Ex14.6 Effect of molecular deformation on polarizability:Page-471 




 
#Variable declaration


# Polarizability of ammonia molecule at 309 K, farad-metre square
alpha_309 = 2.42e-039;                  
# Polarizability of ammonia molecule at 448 K, farad-metre square
alpha_448 = 1.74e-039;                  
k = 1.38e-023;                   # Boltzmann constant, J/mol/K
T1 = 309;                       # First temperature of the experiment, kelvin
T2 = 448;                  # Second temperature of the experiment, kelvin


#Calculation

 # As alpha = alpha_i + alpha_d = alpha_i + p**2/(3*k*T) = alpha_i + bta/T
 # where bta = p**2/(3*k)
 # Thus alpha_309 = alpha_i + bta/309 and alpha_448 = alpha_i + bta/448
 # Solving for bta
 # bta(1/309-1/448) = alpha_309 - alpha_448   
# bta = p**2/(3*k), farad-kelvin metre square
bta= ( alpha_309 - alpha_448)/(1/309 - 1/448);
 # Solving for alpha_i
# Polarizability due to permanent dipole moment, farad-metre square 
alpha_i = alpha_309 - bta/309;   
 # Polarizability due to deformation of molecules = bta/T, bta = p**2/(3*k)
alpha_d_309 = bta/T1# Orientational polarizability at 309 K, farad-metre square
alpha_d_448 = bta/T2#Orientational polarizability at 448 K, farad-metre square



#Result

print"\nThe polarizability due to permanent dipole moment ="
print round((alpha_i*10**40),2)*10**-40," farad-metre square" 
print"\nThe orientational polarization of ammonia at 309 K ="
print round((alpha_d_309*10**39),2)*10**-39,"farad-metre square" 
print"\nThe orientational polarization of ammonia at 448 K ="
print round((alpha_d_448*10**39),2)*10**-39," farad-metre square" 
 
The polarizability due to permanent dipole moment =
2.28e-40  farad-metre square

The orientational polarization of ammonia at 309 K =
2.19e-39 farad-metre square

The orientational polarization of ammonia at 448 K =
1.51e-39  farad-metre square