# Chapter 4 : Insulating Materials¶

## Example 4.1 Page No : 110¶

In [1]:
from sympy import Symbol,solve
# Let C1 and C2 be unknown capacities
#C1+C2 = 0.16
#(C1*C2)/(C1 + C2) = 0.03
# from the above 2 equations we get the following polynomial
s = Symbol("s");
p = s**2 -0.16*s +0.0048

# Calculations
c1 = solve(p)[0]
c2 = 0.16-c1

# Results
print 'Thus the capacitance of condensers is  %3.2f microF  '%(c1)

Thus the capacitance of condensers is  0.04 microF


## Example 4.2 Page No : 110¶

In [2]:
# Variables
n = 9;
Ko = 8.854*10**-12;
K = 5.;
A = 12.*10**-4;
d = 2.*10**-4;

# Calculations
C = (n-1)*Ko*K*A/d

# Results
print 'Thus the capacitance is %e F'%(C);
#The Answer in the Textbook has a calculation error, hence it doesn't match the answer here.

Thus the capacitance is 2.124960e-09 F


## Example 4.3 Page No : 110¶

In [3]:
# Variables
C = 10**-6
V = 10000.

# Calculations
#here C is capacitance and V voltage
E = 1./2*C*V**2
#E is the energy stored in the capacitor
# when the capacitor is discharged all this energy is dissipated as heat in the wire
H = E/4.2

# Results
#H is heat produced in calories math.since 4.2 Joules = 1 calorie
print 'Thus the heat produced is %3.4f calories'%(H)

Thus the heat produced is 11.9048 calories


## Example 4.4 Page No : 111¶

In [4]:
# Variables
A = 0.02;     #surface area of plates in meter square
d = 0.001;     #dismath.tance between the plates in meter
C = 4.5*10**-10;     #capacitance of the capacitor in farad
#for paralel plate condenser C = KoKA/d
Ko = 8.854*10**-12;

# Calculations
#dielectric consmath.tant K is given by
K = (C*d)/(Ko*A)
V = 15000.;     #volatage in volts
Q = C*V     # charge on condenser in columb
D = Q/A     # electric flux density in columb per meter square

# Results
print 'Thus the electric flux density is %e C/m**2)'%(D)

Thus the electric flux density is 3.375000e-04 C/m**2)


## Example 4.5 Page No : 111¶

In [5]:
# Variables
#before inserting the second sheet
d = 0.003;     #distacne between plates in m**2
K1 = 6.;     # relative permittivity of air
Ko = 8.854*10**-12;
# capacitance C1 = Ko*K1*A/d in Farad
#after inserting the second sheet
d1 = 0.003;     #thickness of first sheet in meter
d2 = 0.005;     #thickness of second sheet in meter

# Calculations
#K2 is unknown
#C2 = Ko*A/(d1/K1 + d2/K2)
# but given that C2 = (1/3)*C1
#from equations 1,2,3
K2 =  (d2*K1)/(3*d-d1)
# math.since Ko*A/(d1/K1 + d2/K2) = Ko*K1*A/3*d

# Results
print 'Thus K2 is %3.4f'%(K2)

Thus K2 is 5.0000


## Example 4.6 Page No : 113¶

In [6]:
# Variables
q1 = 1.;     # in coulomb
q2 = 1.;     # in coulomb
Eo = 8.854*10**-12;     # in Farad per meter
Er = 1.;
d = 1.     # in meter
pi = 3.14;

# Calculations
# F is the force between 2 charges in NEWTONS
F = (q1*q2)/(4*pi*Eo*Er*d**2)

# Results
print 'Thus the force between 2 charges is %e'%(F)

Thus the force between 2 charges is 8.992301e+09


## Example 4.7 Page No : 114¶

In [7]:
import math

# Variables
#q1 = q2 = q
pi = 3.14;
d = 0.2;     # in meters
K = 9*10**9;     # here K = 1/4*pi*Eo*Er consmath.tant
F = 9.81*10**-1;     # in newtons or 10**-1 kgm

# Calculations
q = math.sqrt((F*(d**2))/K)

# Results
print 'Thus charge is %e in coulomb'%(q)

Thus charge is 2.088061e-06 in coulomb