Chapter 7:Dielectric Properties

Example 7.1 , Page no:187

In [1]:
import math
from __future__ import division

#given
A=100; #in cm^2 (cross-sectional area)
d=1; #in cm (seperation between plates)
Eo=8.85E-12; #in F/m (absolute permittivity)
V=100; #in V (potential difference)

#calculate
A1=A*1E-4; #changing unit from cm^2 to m^2
d1=d*1E-2; #changing unit from cm to m
C=Eo*A1/d1; #calculation of capacitance
Q=C*V; #calculation of charge
C1=C*1E12; #changing unit of capacitance from F to pF

#result
print"The capacitance of capacitor is C=",C,"C";
print"\t\t\t\t =",C1,"C";
print"The charge on the plates is Q=",Q,"C";

The capacitance of capacitor is C= 8.85e-12 C
				 = 8.85 C
The charge on the plates is Q= 8.85e-10 C

Example 7.2 , Page no:187

In [2]:
import math
from __future__ import division

#given
A=650; #in mm^2 (cross-sectional area)
d=4; #in mm (seperation between plates)
Eo=8.85E-12; #in F/m (absolute permittivity)
Er=3.5; #di-electric constant of the material
Q=2E-10; #in C (charge on plates)

#calculate
A1=A*1E-6; #changing unit from mm^2 to m^2
d1=d*1E-3; #changing unit from mm to m
C=Er*Eo*A1/d1; #calculation of capacitance
V=Q/C; #calculation of charge
C1=C*1E12; #changing unit of capacitance from F to pF

#result
print"The capacitance of capacitor is C=",'%.3E'%C,"C",
print"\n\t\t\t         =",round(C1,2),"pF";
print"The resultant voltage across the capacitor is V=",round(V,2),"V";
print "NOTE: The answer in the textbook is wrong" 

The capacitance of capacitor is C= 5.033E-12 C 
			         = 5.03 pF
The resultant voltage across the capacitor is V= 39.73 V
NOTE: The answer in the textbook is wrong

Example 7.3 , Page no:188

In [3]:
import math
from __future__ import division

#given
N=2.7E25; #in 1/m^3 (density of atoms)
E=1E6; #in V/m (electric field)
Z=2; #atomic number of Helium 
Eo=8.85E-12; #in F/m (absolute permittivity)
Er=1.0000684; #(dielectric constant of the material)
e=1.6E-19; #in C (charge of electron)
pi=3.14; #value of pi used in the solution

#calculate
#since  alpha=Eo*(Er-1)/N=4*pi*Eo*r_0^3 
#Therefore we have r_0^3=(Er-1)/(4*pi*N)
r_0=((Er-1)/(4*pi*N))**(1/3); #calculation of radius of electron cloud
x=4*pi*Eo*E*r_0/(Z*e); #calculation of dispalcement

#result
print"The radius of electron cloud is r_0=",'%.3E'%r_0,"m";
print"The displacement is x=",'%.3E'%x,"m";
print "NOTE: The answer in the textbook is wrong"

The radius of electron cloud is r_0= 5.865E-11 m
The displacement is x= 2.037E+04 m
NOTE: The answer in the textbook is wrong

Example 7.4 , Page no:188

In [4]:
import math
from __future__ import division

#given
K=1.000134; #di-elecrtic constant of the neon gas at NTP
E=90000; #in V/m (electric field)
Eo=8.85E-12; #in C/N-m^2 (absolute premittivity)
N_A=6.023E26; #in atoms/Kg-mole (Avogadro's number)
V=22.4; #in m^3 (volume of gas at NTP

#calculate
n=N_A/V; #calculaton of density of atoms
#Since P=n*p=(k-1)*Eo*E
#therefore we have
p=(K-1)*Eo*E/n; #calculation of dipole moment induced
alpha=p/E; #calculation of atomic polarisability

#result
print"The dipole moment induced in each atom is p=",'%.3E'%p,"C-m";
print"The atomic polarisability of neon is =",'%.3E'%alpha,"c-m^2/V";
print "NOTE: The answer in the textbook is wrong"

The dipole moment induced in each atom is p= 3.969E-36 C-m
The atomic polarisability of neon is = 4.410E-41 c-m^2/V
NOTE: The answer in the textbook is wrong

Example 7.5 , Page no:189

In [5]:
import math
from __future__ import division

#given
Er=3.75; #di-elecrtic constant of sulphur at 27 degree Celcius
gama=1/3; #internal field constant
p=2050; #in Kg/m^3 (density)
M_A=32; #in amu (atomic weight of sulphur)
Eo=8.85E-12; #in F/m (absolute permittivity)
N=6.022E23; #Avogadro's number

#calculate
#Since ((Er-1)/(Er+2))*(M_A/p)=(N/(3*Eo))*alpha_e
#therefore we have
alpha_e=((Er-1)/(Er+2))*(M_A/p)*(3*Eo/N); #calculation of electronic polarisability of sulphur

#result
print"The electronic polarisability of sulphur is =",'%.3E'%alpha_e,"Fm^2   (roundoff error)";

The electronic polarisability of sulphur is = 3.291E-37 Fm^2   (roundoff error)

Example 7.6 , Page no:189

In [6]:
import math
from __future__ import division

#given
Er=1.0000684; #di-elecrtic constant of Helium gas at NTP
Eo=8.85E-12; #in F/m (absolute permittivity)
N=2.7E25; #number of atomsper unit volume

#calculate
#Since Er-1=(N/Eo)*alpha_e
#therefore we have
alpha_e=Eo*(Er-1)/N; #calculation of electronic polarisability of Helium

#result
print"The electronic polarisability of Helium gas is =",'%.3E'%alpha_e,"Fm^2    (roundoff error)";

The electronic polarisability of Helium gas is = 2.242E-41 Fm^2    (roundoff error)

Example 7.7 , Page no:190

In [7]:
import math
from __future__ import division

#given
N=3E28; #in atoms/m^3 (density of atoms)
alpha_e=1E-40; #in F-m^2 (electronic polarisability)
Eo=8.85E-12; #in F/m (absolute permittivity)

#calculate
#Since (Er-1)/(Er+2)=N*alpha_e/(3*Eo)
#therefore we have
Er=(2*(N*alpha_e/(3*Eo))+1)/(1-(N*alpha_e/(3*Eo)));
#calculation of dielectric constant of the material

#result
print"The dielectric constant of the material is Er=",'%.3E'%Er,"F/m",
print "NOTE: The answer in the book is wrong "

The dielectric constant of the material is Er= 1.382E+00 F/m NOTE: The answer in the book is wrong

Example 7.8 , Page no:190

In [8]:
import math
from __future__ import division

#given
Er=4; #relative permittivity of sulphur
Eo=8.85E-12; #in F/m (absolute permittivity)
NA=2.08E3; #in Kg/m^3 (density of atoms in sulphur)

#calculate
#Since ((Er-1)/(Er+2))*(M_A/p)=(N/(3*Eo))*alpha_e
#therefore we have
alpha_e=((Er-1)/(Er+2))*(3*Eo/NA); #calculation of electronic polarisability of sulphur

#result
print"The electronic polarisability of sulphur is =",'%.3E'%alpha_e,"Fm^2";
print "NOTE: The answer in the book is wrong and they used the wrong formula " 

The electronic polarisability of sulphur is = 6.382E-15 Fm^2
NOTE: The answer in the book is wrong and they used the wrong formula

Example 7.9 , Page no:190

In [9]:
import math
from __future__ import division

#given
alpha1=2.5E-39; #in C^2-m/N (dielectric constant at 300K)
alpha2=2.0E-39; #in C^2-m/N (dielectric constant at 400K)
T1=300; #in K(first temperature)
T2=400; #in K(second temperature)

#calculate
#since alpha=alpha_d+alpha0 and alpha0=Beta/T
#therefore alpha=alpha_d+(Beta/T)
#since alpha1=alpha_d+(Beta/T1)  and alpha2=alpha_d+(Beta/T2)
#therefore alpha1-apha2=Beta*((1/T1)-(1/T2))
#or Beta= (alpha1-apha2)/ ((1/T1)-(1/T2))
Beta= (alpha1-alpha2)/ ((1/T1)-(1/T2)); #calculation of Beta
alpha_d=alpha1-(Beta/T1); #calculation of polarisability due to defromation
alpha0_1=Beta/T1; #calculation of polarisability due to permanent dipole moment at 300K
alpha0_2=Beta/T2;  #calculation of polarisability due to permanent dipole moment at 400K

#result
print"The polarisability due to permanent dipole moment at 300K is =",'%.3E'%alpha0_1,"C^2-m/N";
print"The polarisability due to permanent dipole moment at 400K is =",'%.3E'%alpha0_2,"C^2-m/N";
print"The polarisability due to deformation of the molecules is =",'%.3E'%alpha_d,"C^2-m/N";

The polarisability due to permanent dipole moment at 300K is = 2.000E-39 C^2-m/N
The polarisability due to permanent dipole moment at 400K is = 1.500E-39 C^2-m/N
The polarisability due to deformation of the molecules is = 5.000E-40 C^2-m/N

Example 7.10 , Page no:191

In [10]:
import math
from __future__ import division

#given
n=1.5; #refractive index
Er=5.6; #dielectric constant

#calculate
#since (Er-1)/(Er+2)=N*(alpha_e+alpha_i)/(3*E0)   Clausius-Mossotti equation
#and  (n^2-1)/(n^2+2)=N*alpha_e/(3*E0) 
#from above two equations, we get  ((n^2-1)/(n^2+2))*((Er+2)/(Er-1))=alpha_e/(alpha_e+alpha_i)
#or alpha_i/ (alpha_e+alpha_i)= 1-((n^2-1)/(n^2+2))*((Er+2)/(Er-1))= (say P)
#where P is fractional ionisational polarisability
P=1-((n**2-1)/(n**2+2))*((Er+2)/(Er-1)); #calculation of fractional ionisational polarisability
P1=P*100; #calculation of percentage of  ionisational polarisability

#result
print"The percentage of  ionisational polarisability is =",round(P1,1),"percent";

The percentage of  ionisational polarisability is = 51.4 percent