Chapter 3: Electric fields and Capacitors

Example 3.1, Page 79

In [1]:
#Variable declaration
Q = 50e-03;                 # Electric charge, C
A = 600e-06;                # Area of plate, m^2

#Calculations
# Solving for electric field density, D
D = Q/A;                      # Electric field density, C/m^2

#Result
print "The density of the electric field existing between the plates = %4.1f C/m-square"%D

The density of the electric field existing between the plates = 83.3 C/m-square

Example 3.2, Page 80

In [2]:
#Variable declaration
A = 400e-06;              # Cross-sectional area of plate, m^2
I = 50e-06;              # Source current, A
t = 3;                   # Flow time of current, s

#Calculations
# Since electric current is the rate of flow of charge i.e I = Q/t, solving for Q 
Q = I*t;                 # Amount of charge on plates, C
#Solving for density of the electric field between the plates
D = Q/A;                    # Electric field density, C/m^2

#Results
print "The charge on the plates = %3d micro-coloumb"%(Q/1e-06);
print "The density of the electric field between the plates = %5.3f C/m-square"%D

The charge on the plates = 150 micro-coloumb
The density of the electric field between the plates = 0.375 C/m-square

Example 3.3, Page 83

In [3]:
#Variable declaration
d = 3e-03;                  # Thickness of dielectric, m
Q = 35e-03;                 # Electric charge on plates, C
V = 150;                   # Supply voltage, V
A = 144e-06;              # Cross-sectional area of plates, m^2

#Calculations
# Part (a)
# Since electric field strength(E) = potential gradient therefore we have
E = V/d;                    # Electric field strength, V/m
# Part (b)
# Solving for electric field density, D
D = Q/A;                      # Electric field density, C/m^2

#Results
print "The electric field strength = %2d kV/m"%(E*1e-03);
print "The flux density = %5.1f C/m^2"%D

The electric field strength = 50 kV/m
The flux density = 243.1 C/m^2

Example 3.4, Page 83

In [4]:
#Variable declaration
d = 4e-03;                   # Thickness of air, m
Q = 2e-04;                  # Electric charge on plates, C
V = 125;                   # Supply voltage, V
D = 15;                   # Electric field density, coulomb-per-metre-square

#Calculations
# Part (a)
# Since electric field strength(E) = potential gradient, therefore we have 
E = V/d;                    # Electric field strength, V/m
# Part (b)
# Since D = Q/A, solving for A
A = Q/D;              # Cross-sectional area of plates, m^2
# Part (c)
# Since Q = C*V, solving for C
C = Q/V;                    # Capacitance of the plates, F

#Results
print "The electric field strength between the plates = %5.2f kV/m"%(E*1e-03);
print "The csa of the field between the plates = %4.1f mm^2"%(A/1e-06);
print "The capacitance of the plates = %3.1f micro-coulomb"%(C/1e-06);

The electric field strength between the plates = 31.25 kV/m
The csa of the field between the plates = 13.3 mm^2
The capacitance of the plates = 1.6 micro-coulomb

Example 3.5, Page 86

In [5]:
#Variable declaration
A = 6e-04;                          # Cross-sectional area of plates, m^2
d = 5e-04;                          # Thickness of mica sheet, m
Epsilon_r = 5.8;                    # Relative permittivity, unitless
Epsilon_0 = 8.854e-12;              # Permittivity of Free Space
V = 200;                            # Potential difference, V

#Calculations
# Part (a)
# Since absolute permittivity, Epsilon = C*(d/A), therefore solving for d & putting Epsilon = Epsilon_0*Epsilon_r 
C = ( Epsilon_r*Epsilon_0*A )/d;         # Capacitance, F
# Part (b)
# Since electric field strength(E) = potential gradient, therefore we have 
E = V/d;                    # Electric field strength, V/m

#Results
print "The capacitance of the capacitor = %5.2f pF"%(C/1e-12);
print "Electric field strength = %3d kV/m"%(E*1e-03);

The capacitance of the capacitor = 61.62 pF
Electric field strength = 400 kV/m

Example 3.6, Page 86

In [6]:
#Variable declaration
C = 0.224e-09;                    #Capacitance, F
A = 5625e-06;                     # Cross-sectional area of plates, m^2
Epsilon_r = 2.5;                  # Relative permittivity
Epsilon_0 = 8.854e-12;            # Permittivity of Free Space

#Calculations
# Since absolute permittivity, Epsilon = C*(d/A), therefore solving for d & putting Epsilon = Epsilon_0*Epsilon_r
d = ( Epsilon_r*Epsilon_0*A )/C;                  # Thickness of waxed paper dielectric, m

#Result
print "The thickness of paper required = %3.2f mm"%(d/1e-03);

The thickness of paper required = 0.56 mm

Example 3.7, Page 86

In [7]:
#Variable declaration
C = 4.7e-08;                   #Capacitance, F
A = 4e-04;                   # Cross-sectional area of plates, m^2
d = 1e-04;                  # Thickness of dielectric, m
Epsilon_0 = 8.854e-12;            # Permittivity of Free Space

#Calculations
# Since absolute permittivity, Epsilon = C*(d/A), therefore solving for Epsilon_r & putting Epsilon = Epsilon_0*Epsilon_r
Epsilon_r = (C*d)/(Epsilon_0*A);                  # Relative permittivity

#Result
print "Relative permittivity = %4d"%Epsilon_r

Relative permittivity = 1327

Example 3.8, Page 87

In [8]:
#Variable declaration
V = 180;                            # Potential difference, V
d = 3e-03;                         # Thickness of dielectric, m
A = 4.2e-04;                       # Cross-sectional area of plates, m^2
Epsilon_r = 3.5;                   # Relative permittivity
Epsilon_0 = 8.854e-12;             # Permittivity of Free Space

#Calculations
# Since absolute permittivity, Epsilon = C*(d/A), therefore solving for C & putting Epsilon = Epsilon_0*Epsilon_r 
C = ( Epsilon_r*Epsilon_0*A )/d;         # Capacitance, F
# Since C = Q/V, solving for Q
Q = C*V;                                 # Electric charge, C
# Using  D = Q/A,
D = Q/A;                                # Electric field density, C/m^2

#Results
print "The flux thus produced = %3.2f nC."%(Q/1e-09)
print "The flux density thus produced. = %3.2f micro-coulomb-per-metre-square"%(D/1e-06); 
 

The flux thus produced = 0.78 nC.
The flux density thus produced. = 1.86 micro-coulomb-per-metre-square

Example 3.9, Page 89

In [9]:
#Variable declaration
C_1 = 4.7e-06;                   #Capacitance, F
C_2 = 3.9e-06;                   #Capacitance, F
C_3 = 2.2e-06;                   #Capacitance, F

#Calculations
# The resulting capacitance of parallerly connected capacitors is the sum of the individual capacitance present 
#in the circuit
C = C_1 + C_2 + C_3;             # Resulting capacitance  of the circuit, F

#Result
print "The resulting capacitance of the combination = %4.1f micro-farad"%(C/1e-06);

The resulting capacitance of the combination = 10.8 micro-farad

Example 3.10, Page 90

In [10]:
#Variable declaration
C_1 = 6e-06;                    #Capacitance, F
C_2 = 4e-06;                   #Capacitance, F
V = 150;                      # Supply voltage, V

#Calculations
# Part (a)
# The reciprocal of the resulting capacitance of capacitors connected in series is the sum of the reciprocal 
#of the individual capacitances present in the circuit i.e 1/C =  1/C1 + 1/C2, solving for C
C = ( C_1*C_2 )/(C_1 + C_2);               # Resulting capacitance, F
# Part (b)
Q = V*C;                                  # Electric charge on the capacitors, C
# Part (c)
V_1 = Q/C_1;                              # P.d across capacitor C_1, V
V_2 = Q/C_2;                              # P.d across capacitor C_2, V

#Results
print "The total capacitance of the combination = %3.1f micro-farad"%(C/1e-06);
print "The charge on each capacitor = %3d micro-coulomb"%(Q/1e-06);
print "The p.d. developed across %1d micro-farad capacitor = %2d V"%(C_1/1e-06, V_1);
print "The p.d. developed across %1d micro-farad capacitor = %2d V"%(C_2/1e-06, V_2);

The total capacitance of the combination = 2.4 micro-farad
The charge on each capacitor = 360 micro-coulomb
The p.d. developed across 6 micro-farad capacitor = 60 V
The p.d. developed across 4 micro-farad capacitor = 90 V

Example 3.11, Page 91

In [11]:
#Variable declaration
C_1 = 3e-06;                    #Capacitance, F
C_3 = 12e-06;                   #Capacitance, F
C_2 = 6e-06;                    #Capacitance, F
V = 400;                        # Supply voltage, V

#Calculations
# The reciprocal of the resulting capacitance of capacitors connected in series is the sum of the reciprocal of the 
#individual capacitances present in the circuit i.e 1/C =  1/C1 + 1/C2 + 1/C_3, solving for C
C = (C_1 * C_2 * C_3)/( C_1*C_2 + C_2*C_3 + C_3*C_1);               # Resulting capacitance, F
Q = V*C;                                  # Electric charge on the capacitors, C
# Part (c)
V_1 = Q/C_1;                              # P.d across capacitor C_1, V
V_2 = Q/C_2;                              # P.d across capacitor C_2, V
V_3 =Q/C_3;                               # P.d across capacitor C_2, V

#Results
print "P.d across capacitor %1d micro-farad = %5.1f V"%(C_1/1e-06, V_1);
print "P.d across capacitor %1d micro-farad = %5.1f V"%(C_2/1e-06, V_2);
print "P.d across capacitor %2d micro-farad = %4.1f V"%(C_3/1e-06, V_3);

P.d across capacitor 3 micro-farad = 228.6 V
P.d across capacitor 6 micro-farad = 114.3 V
P.d across capacitor 12 micro-farad = 57.1 V

Example 3.12, Page 92

In [12]:
#Variable declaration
V = 200;                      # Supply voltage, voltage
C_AB = 2.;                # Capacitance across branch AB, micro-farad
C_BC = 3.;                 # Capacitance across branch BC, micro-farad
C_CD = 6.;                 # Capacitance across branch CD, micro-farad
C_EF = 8.;                # Capacitance across branch EF, micro-farad
C_BD = 4.;                 # Capacitance across branch EF, micro-farad

#Calculations
# Part (a)
# Since 3-micro-farad & 6-micro-farad  capacitors are in series & the reciprocal of the resulting capacitance of 
#capacitors connected in series is the sum of the reciprocal of the individual capacitances present in the circuit, 
#therefore i.e 1/C =  1/C1 + 1/C2 
C_BCD = ( C_BC*C_CD )/(C_BC+C_CD);         # Resulting capacitance across branch BCD, micro-farad
#Since C_BCD & 4-micro-farad  capacitors are in parallel & the resulting capacitance of parallerly connected capacitors 
#is the sum of the individual capacitance present in the circuit
C_BD = C_BCD + C_BD;         # Resulting capacitance across branch BD, micro-farad
#  Since 2-micro-farad & C_BD  capacitors are in series & the reciprocal of the resulting capacitance of capacitors 
#connected in series is the sum of the reciprocal of the individual capacitances present in the circuit, therefore, 
#we have
C_AD = (C_BD*C_AB)/(C_BD+C_AB);         # Resulting capacitance across branch AD, micro-farad
#Since C_AD & C_EF  capacitors are in parallel & the resulting capacitance of parallerly connected capacitors is the 
#sum of the individual capacitance present in the circuit
C = C_AD + C_EF;         # Resulting capacitance of the circuit, micro-farad
Q = V*C;                                  # Electric  charge drawn from the supply, C

# Part (b)
Q_EF = V*C_EF;                        # The charge on the 8 micro-farad capacitor, micro-coulomb

# Part (c)
Q_AD = Q - Q_EF;                        # The charge on the 4 micro-farad capacitor, C
Q_BD = Q_AD;    # Charge in series combination of capacitors, micro-farad
# Since Q = C*V, solving for V
V_BD = Q_BD/C_BD;                    # The p.d. across the 4F capacitor,V

# Part(d)
Q_BCD = V_BD*C_BCD;                  # Electric charge across branch BCD, C
Q_BC = Q_BCD;                        # Electric charge, C
V_BC = Q_BC/C_BC;                    # The p.d. across the 3 micro-farad capacitor

#Results
print "The charge drawn from the supply = %3.1f mC"%(Q/1e+03);
print "The charge on the %1d micro-farad capacitor = %3.1f mC"%(C_EF, Q_EF/1e+03);
print "The p.d. across the %1d micro-farad capacitor= %2d V"%(C_BD, V_BD);
print "The p.d. across the %1d micro-farad capacitor = %5.2f V"%(Q_BC, V_BC);

The charge drawn from the supply = 1.9 mC
The charge on the 8 micro-farad capacitor = 1.6 mC
The p.d. across the 6 micro-farad capacitor= 50 V
The p.d. across the 100 micro-farad capacitor = 33.33 V

Example 3.13, Page 96

In [13]:
#Variable declaration
N = 20;                  # Number of plates in a capacitor
A = 6400e-06;            # Cross - sectional area of plate, m^2
d = 1.5e-03;             # Distance between plates, m
epsilon_r = 6.4;         # Relative permittivity for mica
epsilon_o = 8.854e-12;     # Relative permittivity for free space

#Calculations
# Calculating the capacitance of the capacitor
C = ((epsilon_o)*(epsilon_r)*A*(N-1))/d;       # Capacitance, F

#Result
print " The capacitance of the capacitor = %3.1f nF"%(C/1e-09);

 The capacitance of the capacitor = 4.6 nF

Example 3.14, Page 96

In [14]:
#Variable declaration
N = 9;                      # Number of plates in a capacitor
A = 1200e-06;               # Cross - sectional area of plate, m^2
C = 3e-10;                  # Capacitance, F
epsilon_r = 5;              # Relative permittivity for mica
epsilon_o = 8.854e-12;      # Relative permittivity for free space

#Calculations
# Using the formula of capacitance, C = ((epsilon_o)*(epsilon_r)*A*(N-1))/d and solving for d, we have
d = ((epsilon_o)*(epsilon_r)*A*(N-1))/C;        # Distance between plates, m

#Result
print "The thickness of mica between parallel plates of a capacitor = %4.2f mm"%(d/1e-03);

The thickness of mica between parallel plates of a capacitor = 1.42 mm

Example 3.15, Page 97

In [20]:
import math

#Variable declaration
N = 11;                       # Number of plates in a capacitor
r = 25e-03;                   # Radius of circular plate, m
A = (math.pi*r**2);                # Cross - sectional area of plate, m^2
d = 5e-04;                    # Distance between plates, m
epsilon_r = 1;                # Relative permittivity for air
epsilon_o = 8.854e-12;        # Relative permittivity for free space

#Calculations
# Calculating the capacitance of the capacitor
C = ((epsilon_o)*(epsilon_r)*A*(N-1))/d;       # Capacitance, F

#Result
print " The capacitance of the capacitor = %3.2f pF"%(C/1e-10);

 The capacitance of the capacitor = 3.48 pF

Example 3.16, Page 99

In [16]:
#Variable declaration
C_1 = 3e-06;                                  # Capacitance, F
C_2 = 6e-06;                                  # Capacitance, F
V_1 = 250;                                   # Voltage across capacitor C_1, V

#Calculations
# Since each capacitor will take charge according to its capacitance, so we have
Q = C_1*V_1;                                 # Charge on first capacitor C_1, C
W_1 = 0.5*C_1*(V_1**2);                       # Energy stored, J
# When the two capacitors are connected in parallel the 3 micro-farad will share its charge with 6 micro-farad capacitor. Thus the total charge in the system will remain unchanged, but the total capacitance will now be different
C = C_1 + C_2;                              # Total capacitance, F
# Since Q = C*V, solving for V
V = Q/C;                                    # Voltage across capacitor C_2, V
W = 0.5*C*(V**2);                            # Total energy stored by the combination, J

#Results
print "The charge and energy stored by %1d micro-F capacitor are %3.2f mC and %5.2f mJ respectively "%(C_1/1e-06, Q/1e-03 , W_1/1e-03);
print "The p.d. between the plates = %5.2f V"%V
print "The energy stored by the combination of %1d micro-F and %1d micro-F capacitors = %5.2f mJ"%(C_1/1e-06, C_2/1e-06, W/1e-03);

The charge and energy stored by 3 micro-F capacitor are 0.75 mC and 93.75 mJ respectively 
The p.d. between the plates = 83.33 V
The energy stored by the combination of 3 micro-F and 6 micro-F capacitors = 31.25 mJ

Example 3.17, Page 100

In [17]:
#Variable declaration
V = 200;                       # Supply voltage, V
C_1 = 10e-06;                  # Capacitance, farad
C_2 = 6.8e-06;                 # Capacitance, farad
C_3 = 4.7e-06;                 # Capacitance, farad

#Calculations
# Part (a)
# Since each capacitor will take charge according to its capacitance, so we have
Q_1 = V*C_1;                   # Charge stored on  capacitor C_1, C
W_1 = 0.5*C_1*(V**2);           # Energy stored on  capacitor C_1, J
# Part (b)
# Since C_2 and C_3 are in series and hence, their equivalent capacitance is given by their series combination
C_4 = (C_2 * C_3)/(C_2 + C_3);          # Equivalent capacitance of C_2 and C_3, F
# Since C_1 and C_4 are in parallel and hence, their equivalent capacitance is given by their parallel combination
C = C_1 + C_4;               # Total capacitance of circuit, F
# Since Q = C*V, solving for V
V_1 = Q_1/C;                         # New p.d across C_1, V
W = 0.5*C*(V_1**2);                 # Total energy remaining in the circuit, J
energy_used = W_1 - W;             # Energy, J

#Results
print "The charge and energy stored by %2d micro-F capacitor are %1d mC and %2.1f J respectively "%(C_1/1e-06, Q_1/1e-03, W_1);
print "The new p.d across %2d micro-F capacitor = %5.1f V"%(C_1/1e-06,V_1);
print "The amount of energy used in charging %3.1f micro-F and %3.2f micro-F capacitors from %2d micro-F capacitor = %4.3f J"%(C_2/1e-06, C_3/1e-06, C_1/1e-06, energy_used/1e-03);
 

The charge and energy stored by 10 micro-F capacitor are 2 mC and 0.2 J respectively 
The new p.d across 10 micro-F capacitor = 156.5 V
The amount of energy used in charging 6.8 micro-F and 4.70 micro-F capacitors from 10 micro-F capacitor = 43.495 J

Example 3.18, Page 101

In [18]:
#Variable declaration
V = 400;               # Supply voltage, V
E = 0.5e06;            # Dielectric strength, V/m

#Calculations
# Since E = V/d, solving for d
d = V/E;               # Thickness of dielectric, m

#Result
print "The minimum thickness of dielectric required = %3.1fmm"%(d/1e-03);

The minimum thickness of dielectric required = 0.8mm

Example 3.19, Page 101

In [19]:
#Variable declaration
C = 270e-12;              # Capacitance, F
A = 60e-04;               # Cross-sectional area of plate, m^2
E = 350e03;               # Dielectric strength, V/m
epsilon_r = 2.1;           # Relative permittivity
epsilon_o = 8.854e-12;     # Permittivity of free space

#Calculations
# Part (a)
# Since formula for capacitance, C = ((epsilon_o)*(eplison_r)*A)/d, solving for d
d = ((epsilon_o)*(epsilon_r)*A)/C;      # Thickness of dielectric, m
# Part (b)
# Since E = V/d, solving for V
V = E*d;                                  # Maximum possible working voltage, V

#Results
print "The thickness of Teflon sheet required = %5.4f mm"%(d/1e-03);
print "The maximum possible working voltage for the capacitor = %5.1f V"%V;

The thickness of Teflon sheet required = 0.4132 mm
The maximum possible working voltage for the capacitor = 144.6 V