# Ch-20, Dielectric, Ferroelectric, piezoelastic & pyroelastic material¶

## example-20.1 page no-593¶

In :
from __future__ import division
#given
#dielecric strength of natural rubber
DS=40000  #volts/mm
#current
I=33*10**3  #V
#required thickness of insulation
t=I/DS  #mm
print "the thickness of wire required for insulation is %0.3f mm"%(t)

the thickness of wire required for insulation is 0.825 mm


## example-20.2 page no-594¶

In :
from math import pi
#given
#capacitance of capacitor
C=0.025*10**-6  #F
#power factor
#tan delta=0.0005
#for notation let tan delta=delta
delta=0.0005
#current
I=200  #A
#frequency
f=25*10**3  #Hz
#volatge across a capacitor is
V=I/(2*(pi)*f*C)  #V
#dielectric loss
P=V*I*delta  # W
print " the dielectric loss is %0.3f Watt"%(P)

 the dielectric loss is 5092.958 Watt


## example-20.3 page no-596¶

In :
#given
#electric field
E=600  #V/m
#dielectric constant
Er=6.1
E0=8.85*10**-12
#polarization is given by
P=E*E0*(Er-1)  #V/m
print "The polarization produced is %0.3e V/m"%(P)

The polarization produced is 2.708e-08 V/m


## example-20.4 page no-596¶

In :
#given
#charge
Q=10*10**-6  #C
#voltage
V=10*10**3  #V
#seperation betweemn the plates
d=5*10**-4  #m
#dielectric eonstant
Er=10
E0=8.854*10**-12
#we know that
#Q=C*V
#so
C=Q/V  #F
#also we know that
#C=Er*E0*A/d
#so
A=C*d/Er/E0  #m**2
print "area between the plates is %0.3f m**2"%(A)

area between the plates is 0.006 m**2


## example-20.5 page no-597¶

In :
#given
#area of plate
A=10*10*10**-6  #m**2
#capacitance
C=10**-9  #F
#distance between the plates
d=2*10**-3  #m
#contant
E0=8.854*10**-12  #F/m
#dielectric constant
Er=C*d/(E0*A)
print "the cielectric constant of the crystal is %0.3f"%(Er)

the cielectric constant of the crystal is 2258.866


## example-20.6 page no-597¶

In :
#given
#dielectric constant
Er1=6.0
Er2=3.0
#thickness of plates
d1=0.25*10**-3  #m
d2=0.1*10**-3  #m
#taking A1=A2
#we know that
#C=Er*E0*A/d
#for plate1
#C1=Er1*E0*A1/d1     -----------(1)
#for plate 2
#C2=Er1*E0*A2/d2   --------------(2)
#dividing 1 and 2
#we get
#C1/C2=Er1*d2/(Er2*d1)
#let C1/C2=c
C=Er1*d2/(Er2*d1)
#so we get
#C1=0.8*C2
print "The plastic film wil hold more charge"

The plastic film wil hold more charge


## example-20.7 page no-609¶

In :
#given
#thickness of BaTiO3 wafer
t=0.15*10**-3  #m
#compressive strength
sigma=25*10**6  #N/m**2
#young's modulus of elasticity
Y=70*10**9  #N/m**2
#electric field E produced by the stress sigma is related as
#E=sigma*lambda  where lambda is constant known as voltage output coefficient
lamda=1*10**-10  #m/V
#and modulus of elasticity is gven by
#Y=1/(lambda*t)
#so we get from 1 and 2
#E=sigma/(Y*t)
#also E=V/t
#so
V=sigma*t/(lamda*Y)  #V
print "Potential difference producd across tha wafer is %0.3f V"%(V)

Potential difference producd across tha wafer is 535.714 V


## example-20.8 page no-613¶

In :
from math import sqrt
#given
#vibration frequency
f=434*10**3 #Hz
#young's modulus of elasticity
E=80*10**9   #Pa
#density
rho=2655  #kg/m**3
#and fundamental overtones may be 1,2,3.......
#so
n=1
#we konow tha
#f=n/(2*t)*sqrt(E/rho) wher t is the thickness of crystal
#so
t=n/(2*f)*sqrt(E/rho)*10**3  #mm
print " the thickness of crystal is %0.3f mm"%(t)

 the thickness of crystal is 6.324 mm


## example-20.9 page no-613¶

In :
#given
#capacitance of paper capacitance
C=0.02*10**-6  #F
#thickness of capacitor
d=1*10**-3  #m
#relative permitivity
Er=2.6
E0=8.85*10**-12
#dielectric strength
k=1.8*10**7   #V/m
#area of capacitor is given by
A=C*d/(Er*E0)*10**4  #cm**2
#breakdown volatage
Vbreakdown=k*d  #V
print "The area of capacitor is %0.3f am**2 \n and breakdown volatage is %0.3f V"%(A,Vbreakdown)

The area of capacitor is 8691.873 am**2
and breakdown volatage is 18000.000 V


## example-20.10 page no-614¶

In :
#given
#capacitance of capacitor
C=0.2*10**-6  #F
#loss factor
#tan delta=0.004
# for notation let us use tan delta=delta
delta=0.004
#voltage
V=240  #V
#frequency
f=50  #Hz
#and
omega=2*(pi)*f
#power loss is given by
P=V**2*omega*C*delta  #W
print " power loss in the capacitor is %0.3f W"%(P)

 power loss in the capacitor is 0.014 W