# Chapter 5 : Dielectric Materials¶

## Example 5.1 Page No : 193¶

In [1]:
# Variables
t = 0.25     #time in sec
I = 0.22     #Current in A
V = 220.     #voltage in V

# Calculations
Q = I*t     #charge given to condenser
C = Q/V     #capacitance of condenser
C1 = C*(10**6)

# Results
print 'Charge given to condenser is %3.3f Coulombs '%(Q)
print 'Capacitance of condenser is %3.4f F'%(C)
print 'or %3.0f microF'%(C1)

Charge given to condenser is 0.055 Coulombs
Capacitance of condenser is 0.0003 F
or 250 microF


## Example 5.2 Page No : 193¶

In [2]:
# Variables
C = 0.0002*(10**-6)     #capacitance in F
V = 20000.     #P.D across condenser in V
t = 2     #thickness in mm

# Calculations
Q = C*V     #charge on each plate in coulomb
g = (V/t)*(1./1000)     # potential gradient in kV/mm

# Results
print 'Charge given to condenser is %e Coulombs '%(Q)
print 'Potential gradient of condenser is %3.0f kV/mm'%(g)

Charge given to condenser is 4.000000e-06 Coulombs
Potential gradient of condenser is  10 kV/mm


## Example 5.3 Page No : 194¶

In [4]:
# Variables
#Before immersion of oil
C = 0.005*(10**-6)
V = 500.
#After immersion of oil
K = 2.5

# Calculations
q = C*V
q1 = q     # math.since no loss of charge
E = (1./2)*(C*V*V)
C1 = K*C     #capacity of condenser
E1 = (q1**2)/(2*C1)     # energy stored in condenser

# Results
print 'Charge of condenser is %e coulomb '%(q)
print 'Energy stored in condenser before immersion of oil is %e Joules '%(E)
print 'Energy stored in condenser after immersion of oil is %e Joules'%(E1)

Charge of condenser is 2.500000e-06 coulomb
Energy stored in condenser before immersion of oil is 6.250000e-04 Joules
Energy stored in condenser after immersion of oil is 2.500000e-04 Joules


## Example 5.4 Page No : 194¶

In [5]:
# Variables
A = 0.02     #surface area of plate in m**2
d = 0.001     #dismath.tance between plates in m
C = 4.5*(10**-10)     #capacitance in F
V = 15000.     #voltage in volts

# Calculations
K0 = 8.854*(10**-12)
K = (C*d)/(K0*A)
q = C*V     # charge on condenser in coulombs
D = q/A     #Electric flux density in Coulomb/m**2

# Results
print 'Thus dielectric consmath.tant is %3.2f '%(K)
print 'Thus Electric flux density is %e Coulombs/m**2'%(D)

Thus dielectric consmath.tant is 2.54
Thus Electric flux density is 3.375000e-04 Coulombs/m**2


## Example 5.5 Page No : 195¶

In [6]:
# Variables
A = 0.2     #surface area of plate in m**2
t = 2.5*(10**-5)     #thickness of dielectric in m

# Calculations
K0 = 8.854*(10**-12)     #permittivity of air  in F/m
K = 5     #relative permittivity of dielectric
C = (K*K0*A*(10**6))/t     #capacitance of condenser in microF

# Results
print 'Thus the Capacitance of condenser is %3.3f microF'%(C)

Thus the Capacitance of condenser is 0.354 microF