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
```

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
```

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
```

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);
```

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);
```

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);
```

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
```

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);
```

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);
```

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);
```

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);
```

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);
```

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);
```

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);
```

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);
```

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);
```

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);
```

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);
```

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;
```