Chapter 18: Dielectric Materials

Example 18.1, page no-460

In [1]:
# Relative permitivity of KCl

import math
# Variable declaration
atom=4                             # number of atoms             
kci=0.629*10**-9                   # LAttice parameter of KCl   
alfk=1.264*10**-40                 # electronic polarisability for K+ ion
alfCl=3.408*10**-40                # electronic polarisability for Cl- ion
eps0=8.854*10**-12                 # permitivity of free space

# Calculations
pol=alfk+alfCl
N=atom/kci**3
epsr=(N*pol/eps0)+1

#Result
print("\nThe electronic polarisability for KCL = %.3f *10^-40 F m^2\n"%(pol*10**40))
print("\nThe no of Dipoles per m^3 = %.3f * 10^28 atoms m^-3\n"%(N/10**28))
print("\nThe dielectric constant of KCL is %.3f"%epsr)

The electronic polarisability for KCL = 4.672 *10^-40 F m^2


The no of Dipoles per m^3 = 1.607 * 10^28 atoms m^-3


The dielectric constant of KCL is 1.848

Example 18.2, page no-460

In [2]:
# electronic polarisability

import math
#variable declarations
r=0.12*10**-9                  # atomic radius of Se
eps=8.854*10**-12              # permitivity of free space 

# Calculations
alf=4*math.pi*eps*r**3

# Result
print("The electronic polarisability of an isolated Se is %.4f * 10^-40 F m^2"%(alf*10**40))

The electronic polarisability of an isolated Se is 1.9226 * 10^-40 F m^2

Example 18.3, page no-461

In [3]:
#electronic to ionic polarability ratio

import math
# Variable declaration
n=2.69                         # refraction index
er=4.94                        # dielectric cnstant

# calculations
alfi_by_alfe=(((n+2)*(er-1))/((er+2)*(n-1)))-1

# Result
print("The ratio of the electronic to ionic polarability is %.4f"%(1/alfi_by_alfe))

The ratio of the electronic to ionic polarability is 1.7376

Example 18.4, page no-462

In [4]:
# dielectric constant

import math
#variable declaration
N= 2.7*10**25                # number of atoms
alfe=0.35*10**-40            # electronic polarisability 
eps=8.854*10**-12            # permitivity of free space

# calculations
epsr=(1+(2*N*alfe)/(3*eps))/(1-(N*alfe)/(3*eps))

# Result
print("The dielectric constant of Ne gas is %.8f"%epsr)

The dielectric constant of Ne gas is 1.00010674

Example 18.5, page no-462

In [5]:
# charge on the capacitor

import math
# Variable declaration
eps=8.85*10**-12               # permitivity of free space
epsr=6                         # relative permitivity of dielectric
A=5*10**-4                     # Area of the capacitor plate  
d=1.5*10**-3                   # distance between the plates
v=100                          # Applied voltage

# calculations
Q=eps*epsr*A*v/d

# Result
print("The charge on the capacitor is %.2f * 10^-9 C"%(Q*10**9))

The charge on the capacitor is 1.77 * 10^-9 C

Example 18.6, page no-463

In [13]:
# dielectric constant

import math
# variable declaration
N=2.7*10**25                    # Number of Ar atoms
d=0.384*10**-9                  # diameter of Ar atom
eps=8.854*10**-12               # permitivity of free space

# calculations
alfe=4*math.pi*eps*d**3
alfe=alfe*10**-2                # correction
epsr=(1+((2*N*alfe)/(3*eps)))/(1-((N*alfe)/(3*eps)))

# Result
print("The dielectric constant of Ar  is %.8f"%(epsr))
# correction is to match the answer in the book 
# answer for alfe is given as 0.63 * 10^-40 but it is actually 0.63* 10^-38.

The dielectric constant of Ar  is 1.00019213

Example 18.7, page no-464

In [14]:
# Energy stored

import math
# Variable declaration
c=2*10**-6                     # capacitance
epsr=80                        # permitivity of the dielectric
v=1000                         # Applied voltage

#Calculations
E1=(c*v**2)/2
c0=c/epsr
E2=(c0*v**2)/2
E=E1-E2

# Result
print("\nThe Energy stored in capacitor =%.0f J"%E1)
print("\nThe energy stored in polarising the capacitor = %.4f J"%E)

The Energy stored in capacitor =1 J

The energy stored in polarising the capacitor = 0.9875 J

Example 18.8, page no-464

In [15]:
# nternal field to applied field ratio

import math
# Variable declaration
N=5*10**28                  # no of atoms present per m^3
alfe=2*10**-40              # Polarisability    
eps=8.854*10**-12           # permitivity of free space


# Calculations
P=N*alfe
E_ratio=1/(1-(P/(3*eps)))

# Result
print("The ratio of the internal field to the applied field = %.4f"%E_ratio)

The ratio of the internal field to the applied field = 1.6038

Example 18.9, page no-465

In [16]:
# relative permitivity

import math
# Variable declaration
E=1000                           # Applied electric field
P=4.3*10**-8                     # Polarisation
eps=8.854*10**-12                # permitivity of free space

# calculations
epsr=1+P/(eps*E)

#Result
print("The relative permitivity of NaCl is %.2f"%epsr)

The relative permitivity of NaCl is 5.86

Example 18.10, page no-466

In [17]:
#polarisability of argon atom

import math
# Variable declaration
epsr=1.0024                   # relative permitivity
N=2.7*10**25                  # Number of atoms
eps=8.854*10**-12             # permitivity of free space

# calculations
alfe=eps*(epsr-1)/N

#Result
print("The polarisability of argon atom is %.1f * 10^-40 F m^2"%(alfe*10**40))

The polarisability of argon atom is 7.9 * 10^-40 F m^2

Example 18.11, page no-466

In [18]:
# electronic polarisability

import math
# Variable declaration
epsr=1.0000684                     # Dielectric constant of the gas at NTP
N=2.7*10**25                       # Number of He atoms
eps=8.854*10**-12                  # permitivity of free space

#calculations
alfe=eps*(epsr-1)/N

#Result
print("The electronic polarisability of He atom at NTP is %.3f * 10^-41 F m^2"%(alfe*10**41))

The electronic polarisability of He atom at NTP is 2.243 * 10^-41 F m^2

Example 18.12, page no-467

In [22]:
# electronic polarisability

import math
# variable declaration
epsr=12                      # relative dielectric constant of material
N=5*10**28                   # number of atoms in the element
eps=8.854*10**-12            # permitivity of free space

#Calculations
alfe=eps*(epsr-1)/N

# result
print("The electronic polarisability of given element is %.3f * 10^-39 F m^2"%(math.floor(alfe*10**39*1000)/1000))

The electronic polarisability of given element is 1.947 * 10^-39 F m^2

Example 18.13, page no-467

In [23]:
# energy stored in dielectric

import math
# variable declaration
c=2*10**-6                    # capacitance of plate condenser
v=1000                        # applied voltage
epsr=100                      # dielectric permitivity

# calculations
E=(c*v**2)/2
c0=c/epsr
e2=(c0*v**2)/2
E1=E-e2

# Result
print("The energy stored in dielectric is %.2f J"%E1)

The energy stored in dielectric is 0.99 J

Example 18.14, page no-468

In [43]:
# electronic polarisability

import math
#variable declaration
epsr=3.4                        # dielectric constant of sulphur 
eps=8.854*10**-12               # permitivity of free space
d=2.07*10**3                    # density of sulphur 
w=32.07                         # Atomic weight
Avg=6.023*10**23                # avogadro's number

# calculations
N=Avg*10**3*d/w
N= (math.ceil(N*10**-26))/10**-26
alfe=3*eps*(epsr-1)/(N*(epsr+2))

#Result
print("The electronic polarisability of sulphur is %.3f * 10^-40 F.m^2"%(alfe*10**40))

The electronic polarisability of sulphur is 3.035 * 10^-40 F.m^2

Example 18.15, page no-469

In [53]:
# charge stored and polarisation produced in the plate

import math
#variable declaration
A=6.45*10**-4                  # Area of the capacitor plate
d=2*10**-3                     # distance between plates
epsr=6                         # relative permitivity 
v=10                           # applied voltage 
eps=8.854*10**-12              # permitivity in free space


# calculations
c=eps*epsr*A/d
q=c*v
E=v/d
p=eps*(epsr-1)*E

# Result
print("Capacitance of Capacitor = %.2f pF"%(c*10**12))
print("\ncharge stored on the plate is %.2f *10^-11 C"%(q*10**11))
print("\nPolarisation produce in the plate is %.3f *10^-7 Cm^-2"%(math.ceil(p*10**7*1000)/1000))
# answer forstored charge is wrong in the book

Capacitance of Capacitor = 17.13 pF

charge stored on the plate is 17.13 *10^-11 C

Polarisation produce in the plate is 2.214 *10^-7 Cm^-2

Example 18.16, page no-470

In [54]:
# Polarisation produced in NaCl\

import math
# Variable declaration
E=600*10**3                      # electric field strength
eps=8.854*10**-12                # permitivity in free space   
epsr=6                           # dielectric constant of sodium chloride

# calculations
p=eps*(epsr-1)*E

# Result
print("Polarisation produced in NaCl is %.3f *10^-5 C.m^-2"%(p*10**5))

Polarisation produced in NaCl is 2.656 *10^-5 C.m^-2

Example 18.17, page no-470

In [55]:
# Relative permitivity

import math
# Variable declaration
E=1000                      # applied electric field
p=4.3*10**-8                # Polarisation   
eps=8.854*10**-12           # permitivity of free space

#Calculations
epsr=1+p/(eps*E)

#Result
print("Relative permitivity of NaCl is %.2f"%epsr)

Relative permitivity of NaCl is 5.86

Example 18.18, page no-471

In [57]:
# voltage across capacitor and electric field strength

import math
# variable declaration
A=1000*10**-6               # Area of the capacitor plate
d=5*10**-3                  # distance between the plate
epsr=4                      # relative permitivity of the dielectric
Q=3*10**-10                 #charge on the capacitor 
eps=8.854*10**-12           # permitivity of free space

# Calculations
c=(eps*epsr*A)/d
v=Q/c
E=v/d

# Result
print("The voltage across capacitor is %.2f V\nThe electric field strength is %d V/m"%(v,E))

The voltage across capacitor is 42.35 V
The electric field strength is 8470 V/m

Example 18.19, page no-472

In [61]:
# electronic polarisability

import math
# Variable declaration
epsr=1.0000684                    # dielectric constant of the gas at NTP
N=2.7*10**25                      # Number of He atoms 
eps=8.85*10**-12                  # permitivuty of free space

# calculations
alfe=eps*(epsr-1)/N

# Result
print("The electronic polarisability of He atoms at NTP is %.3f *10^-41 F.m^2"%(alfe*10**41))

The electronic polarisability of He atoms at NTP is 2.242 *10^-41 F.m^2

Example 18.20, page no-472

In [71]:
# Capacitance and electric field strength

import math
# Variable declaration
A=3*10**-3                     # area of the capacitor plate
d=1*10**-3                     # distance between the plate
epsr=3.5                       # relative permitivity of the dielectric
Q=20*10**-9                    # charge on the capacitor
eps=8.85*10**-12              # permitivity of free space

# calculations
c=eps*epsr*A/d
E=Q/(c*d)

# Result
print("The capacitance of capacitor is %.2f pF"%(math.ceil(c*10**14)/100))
print("The electric field strength is %.2f*10^3 V/m"%(math.floor(E*10**-1)/100))

The capacitance of capacitor is 92.93 pF
The electric field strength is 215.22*10^3 V/m

Example 18.21, page no-473

In [77]:
# capacitance, stored charge, polarisation and dielectric displacement

import math
# variable declaration
A=7.45*10**-4                        # Area of the capacitor plates
d=2.45*10**-3                        # distance between the plates
epsr=6                               # relative permitivity of the dielectric 
v=10                                 # applied voltage 
eps=8.85*10**-12                    # permitivity of free space

#Calculations
c=eps*epsr*A/d
Q=c*v
E=v/d
p=eps*(epsr-1)*E
D=eps*epsr*E

# Result
print("\nThe capacitance of the capacitor is %.3f pF"%(c*10**12))
print("\nCharge stored on capacitor = %.3f *10^-11 C\n\nE=%.2f*10^3 V/m"%(Q*10**11,E*10**-3))
print("\nPolarisation=%.3f*10^-7 C.m^-2\n\ndielectric displacement = %.3f*10^-7 C.m^-2"%(p*10**7,D*10**7))

The capacitance of the capacitor is 16.147 pF

Charge stored on capacitor = 16.147 *10^-11 C

E=4.08*10^3 V/m

Polarisation=1.806*10^-7 C.m^-2

dielectric displacement = 2.167*10^-7 C.m^-2

Example 18.22, page no-475

In [91]:
# polarisation produced

import math
#variable declaration
E=500                          # electric field strength
epsr=6                         # dielectric constant of sodium cloride
eps=8.854*10**-12              # permitivity of free space

# calculations
p=eps*(epsr-1)*E

# Result
print("The polarisation produced in NaCl is %.3f * 10^-8 C.m^-2"%(math.ceil(p*10**11)/1000))

The polarisation produced in NaCl is 2.214 * 10^-8 C.m^-2

Example 18.23, page no-475

In [92]:
# polarisation produced in NaCl

import math
#Variable declaration
E=500                           # electric field strength
epsr=15                         # dielectric constant of sodium cloride       
eps=8.854*10**-12               # permitivity of free space

#calculations
p=eps*(epsr-1)*E

#Result
print("The polarisation produced in NaCl is %.3f * 10^-8 C.m^-2"%(p*10**8))

The polarisation produced in NaCl is 6.198 * 10^-8 C.m^-2

Example 18.24, page no-475

In [97]:
# Voltage across capacitor

import math
#variable declaration
A=650*10**-6                              # Area of the capacitor plate          
d=4 *10**-3                               # distance between the plates
epsr=3.5                                  # relative permitivity of the dielectric
eps=8.85*10**-12                          # permitivity of free space 
q=2*10**-10                               # charge on the capacitor

# calculations
v=q*d/(eps*epsr*A)

# Result
print("The voltage across capacitor is %.2f V"%v)

The voltage across capacitor is 39.73 V

Example 18.25, page no-476

In [98]:
# charge on the capacitor

import math
# variable declaration
A=5*10**-4                           # Area of the capacitor plates
d=1.5*10**-3                         # Distance between the plates
epsr=6                               # Relative permitivity of the dielectric
v=100                                # Applied voltage
eps=8.854*10**-12                    # permitivity of free space

#calculation
q=eps*epsr*A*v/d

#Result
print("The charge on the capacitor is %.2f *10^-9 C"%(q*10**9))

The charge on the capacitor is 1.77 *10^-9 C

Example 18.26, page no-476

In [3]:
#dielectric constant

import math
#variable declaration
d=2.08*10**3                 # density of sulphur
wt=32                        # atomic weight of sulphur
ep=3.28*10**-40              # electronic polarisability
eps=8.854*10**-15            # permeability of free space

# Calculations
k=(3*10**28*7*10**-40)/(3*eps)
epsr=2.5812/(1-0.7906)

# result
print("The dielectric constant of the given material is %.3f"%epsr)

The dielectric constant of the given material is 12.327