## How to Calculate and Solve for Average Repeat Unit Molecular Weight for Co-Polymers | Polymer

The average repeat unit molecular weight for co-polymers is represented by the image below.

To compute for average repeat unit molecular weight for co-polymers, one essential parameter is needed and the parameter is Numbers of co-polymers (n). By determining the number of co-polymers, four parameters is needed to get your answer so the parameters are Mole Fraction (f1), Molecular Weight (M1), Mole Fraction (f2) and Molecular Weight (M2).

The formula for calculating the average repeat unit molecular weight for co-polymers:

m’ = ΣfiMi

Let’s solve an example;
Find the average repeat unit molecular weight for co-polymers when the mole fraction(s) is 8,2 and the molecular weight(s) is 4,6.

This implies that;

m’ = ΣfiMi

Where:

m’ = Average Repeat Unit Molecular Weight for Co-Polymers
fi = Mole Fraction(s)
Mi = Molecular Weight(s)

m’ = (8)(4) + (2)(6)
m’ = (32) + (12)
m’ = 44

Therefore, the average repeat unit molecular weight for co-polymers is 44.

## How to Calculate and Solve for Degree of Polymerization | Polymer

The image of degree of polymerization is represented above.

To compute for degree of polymerization, two essential parameters are needed and these parameters are Mean Number Molecular Weight (M’nand Unit Molecular Weight (m).

The formula for calculating degree of polymerization:

D.P. = M’n/m

Where:

D.P. = Degree of Polymerization
M’n = Mean Number Molecular Weight
m = Unit Molecular Weight

Let’s solve an example;
Find the degree of polymerization when the mean number molecular weight is 28 and the unit molecular weight is 4.

This implies that;

M’n = Mean Number Molecular Weight = 28
m = Unit Molecular Weight = 4

D.P. = 28/4
D.P. = 7

Therefore, the degree of polymerization is 7.

Calculating the Mean Number Molecular Weight when the Degree of Polymerization and the Unit Molecular Weight is Given.

M’n = D.P (m)

Where;

M’n = Mean Number Molecular Weight
D.P. = Degree of Polymerization
m = Unit Molecular Weight

Let’s solve an example;
Find the mean number molecular weight when the degree of polymerization is 6 and the unit molecular weight is 4.

This implies that;

D.P. = Degree of Polymerization = 6
m = Unit Molecular Weight = 4

M’n = D.P (m)
M’n = 6 (4)
M’n = 24

Therefore, the degree of polymerization is 24.

## How to Calculate and Solve for Number Average Molecular Weight | Polymer

The image of mean is represented below

To compute for mean, two essential parameters are needed and these parameters are Set of data and Corresponding frequencies accordingly.

The formula for calculating mean:

Mean = ∑fx / ∑f

Where;

∑fx = Summation of the product of the data and its corresponding frequency.
∑f = Summation of the frequencies

Let’s solve an example;
Find the mean when the set of data is 12 and the corresponding frequencies accordingly is 2.

This implies that;

x f (fx) fx
12 2 (12 x 2) 24

∑fx = 24
∑f = 2

Mean = ∑fx / ∑f
Mean = 24 / 2
Mean = 12

Therefore, the mean is 12. Continue reading How to Calculate and Solve for Number Average Molecular Weight | Polymer

## How to Calculate and Solve for Concentration Polarization | Corrosion

The image of concentration polarization is shown below.

To compute for concentration polarization, six essential parameters are needed and these parameters are Gas Constant (R), Temperature (T), Number of Electrons (n), Faraday’s Constant (F), Current Density (i) and Limiting Diffusion Current Density (iL).

The formula for calculating concentration polarization:

ηc = 2.3RT/nF log(1 – i/iL)

Where:

ηc = Concentration Polarization
R = Gas Constant
T = Temperature
n = Number of Electrons
F = Faraday’s Constant
i = Current Density
iL = Limiting Diffusion Current Density

Let’s solve an example;
Find the concentration Polarization when the gas constant is 14, the temperature is 7, the number of electrons is 8, the faraday’s constant is 4, the current density is 1 and the limiting diffusion current density is 2.

This implies that;

R = Gas Constant = 14
T = Temperature = 7
n = Number of Electrons = 8
F = Faraday’s Constant = 4
i = Current Density = 1
iL = Limiting Diffusion Current Density = 2

ηc = 2.3RT/nF log(1 – i/iL)
ηc = 2.3(14)(7)/(8)(4) log(1 – (1/2))
ηc = 225.39/32 log(1 – 0.5)
ηc = 7.043749 log(0.5)
ηc = 7.043749 (-0.301029)
ηc = -2.120

Therefore, the concentration polarization is -2.120.

## How to Calculate and Solve for Total Polarization | Electrical Properties

The image of total polarization is shown above.

To compute for total polarization, three essential parameters are needed and these parameters are Electronic Polarization (Pe), Ionic Polarization (Pi) and Orientation Polarization (Po).

The formula for calculating total polarization:

P = Pe + Pi + Po

Where:

P = Total Polarization
Pe = Electronic Polarization
Pi = Ionic Polarization
Po = Orientation Polarization

Let’s solve an example;
Find the total polarization when the electronic polarization is 24, the ionic polarization is 4 and the orientation polarization is 2.

This implies that;

Pe = Electronic Polarization = 24
Pi = Ionic Polarization = 4
Po = Orientation Polarization = 2

P = Pe + Pi + Po
P = 24 + 4 + 2
P = 30

Therefore, the total polarization is 30 C/m².

Calculating the Electronic Polarization when the Total Polarization, the Ionic Polarization and the Orientation Polarization is Given.

Pe = P – Pi – Po

Where:

Pe = Electronic Polarization
P = Total Polarization
Pi = Ionic Polarization
Po = Orientation Polarization

Given an example;
Find the electronic polarization when the total polarization is 29, the ionic polarization is 10 and the orientation polarization is 3.

This implies that;

P = Total Polarization = 29
Pi = Ionic Polarization = 10
Po = Orientation Polarization = 3

Pe = P – Pi – Po
Pe = 29 – 10 – 3
Pe = 19 – 3
Pe = 16

Therefore, the electronic polarization is 16.

## How to Calculate and Solve for Polarization of Dielectric Medium | Electrical Properties

The image of polarization of dielectric medium is shown below.

To compute for polarization of dielectric medium, three essential parameters are needed and these parameters are Permittivity of Vacuum (εo), Dielectric Constant (εr) and Electric Field Intensity (E).

The formula for calculating polarization of dielectric medium:

P = εor – 1)E

Where:

P = Polarization of a Dielectric Medium
εo = Permittivity of Vacuum
εr = Dielectric Constant
E = Electric Field Intensity

Let’s solve an example;
Find the polarization of dielectric medium when the permittivity of vacuum is 12, the dielectric constant is 6 and the electric field intensity is 4.

This implies that;

εo = Permittivity of Vacuum = 12
εr = Dielectric Constant = 6
E = Electric Field Intensity = 4

P = εor – 1)E
P = (12)(6 – 1)(4)
P = (12)(5)(4)
P = 240

Therefore, the polarization of dielectric medium is 240 C/m².

## How to Calculate and Solve for Dielectric Displacement in Relation to Polarization | Electrical Properties

The image of dielectric displacement in relation to polarization is illustrated below.

To compute for dielectric displacement in relation to polarization, three essential parameters are needed and these parameters are Permittivity of Vacuum (εo), Electric Field Intensity (E) and Polarization (P).

The formula for calculating dielectric displacement in relation to polarization:

D = εoE + P

Where:

D = Dielectric Displacement in Relation to Polarization
εo = Permittivity of Vacuum
E = Electric Field Intensity
P = Polarization

Let’s solve an example;
Find the dielectric displacement in relation to polarization when the permittivity of vacuum is 8, the electric field intensity is 6 and the polarization is 2.

This implies that;

εo = Permittivity of Vacuum = 8
E = Electric Field Intensity = 6
P = Polarization = 2

D = εoE + P
D = (8)(6) + 2
D = (48) + 2
D = 50

Therefore, the dielectric displacement in relation to polarization is 50 C/m².

Calculating the Permittivity of Vacuum when the Dielectric Displacement in Relation to Polarization, the Electric Field Intensity and the Polarization is Given.

εo = D – P / E

Where:

εo = Permittivity of Vacuum
D = Dielectric Displacement in relation to polarization
E = Electric Field Intensity
P = Polarization

Let’s solve an example;
Find the permittivity of vacuum when the dielectric displacement in relation to polarization is 22, the electric field intensity is 6 and the polarization is 4.

This implies that;

D = Dielectric Displacement in relation to polarization = 22
E = Electric Field Intensity = 6
P = Polarization = 4

εo = D – P / E
εo = 22 – 4 / 6
εo = 18 / 6
εo = 3

Therefore, the permittivity of vacuum is 3.

## How to Calculate and Solve for Dielectric Displacement | Electrical Properties

The image of dielectric displacement is shown above.

To compute for dielectric displacement, two essential parameters are needed and these parameters are Permittivity of Vacuum (εo) and Electric Field Intensity (E).

The formula for calculating the dielectric displacement;

Do = εoE

Where:

Do = Dielectric Displacement
εo = Permittivity of Vacuum
E = Electric Field Intensity

Given an example;
Find the dielectric displacement when the permittivity of vacuum is 28 and the electric field intensity is 4.

This implies that;

εo = Permittivity of Vacuum = 28
E = Electric Field Intensity = 4

Do = εoE
Do = (28)(4)
Do = 112

Therefore, the dielectric displacement is 112 C/m².

Calculating the Permittivity of Vacuum when the Dielectric Displacement and the Electric Field Intensity is Given.

εo = DoE

Where:

εo = Permittivity of Vacuum
Do = Dielectric Displacement
E = Electric Field Intensity

Let’s solve an example;
Find the permittivity of vacuum when the dielectric displacement is 4 and the electric field intensity is 6.

This implies that;

Do = Dielectric Displacement = 4
E = Electric Field Intensity = 6

εo = DoE
εo = 4(6)
εo = 24

Therefore, the permittivity of vacuum is 24.

## How to Calculate and Solve for Electric Dipole Moment | Electrical Properties

The image of electric dipole moment is shown above.

To compute for electric dipole moment, two essential parameters are needed and these parameters are Magnitude of Each Dipole (q) and Distance of Separation (d).

The formula for calculating electric dipole moment:

p = qd

Where:

p = Electric Dipole Moment
q = Magnitude of Each Dipole
d = Distance of Separation

Let’s solve an example;
Find the electric dipole moment when the magnitude of each dipole is 21 and the distance of seperation is 7.

This implies that;

q = Magnitude of Each Dipole = 21
d = Distance of Separation = 7

p = qd
p = (21)(7)
p = 147

Therefore, the electric dipole moment is 147 C-m.

Calculating the Magnitude of Each Dipole when the Electric Dipole Moment and the Distance of Separation is Given.

q = p / d

Where:

q = Magnitude of Each Dipole
p = Electric Dipole Moment
d = Distance of Separation

Let’s solve an example;
Given that, the electric dipole moment is 12 and the distance of seperation is 4. Find the magnitude of each dipole?

This implies that;

p = Electric Dipole Moment = 12
d = Distance of Separation = 4

q = p / d
q = 12 / 4
q = 3

Therefore, the magnitude of each dipole is 3.

## How to Calculate and Solve for Dielectric Constant | Electrical Properties

The dielectric constant is illustrated by the image below.

To compute for dielectric constant, two essential parameters are needed and these parameters are Permittivity of Dielectric Material (ε) and Permittivity of Vacuum (εo).

The formula for calculating the dielectric constant:

εr = ε/εo

Where:

εr = Dielectric Constant
ε = Permittivity of Dielectric Material
εo = Permittivity of Vacuum

Given an example;
Find the dielectric constant when the permittivity of dielectric material is 36 and the permittivity of vacuum is 9.

This implies that;

ε = Permittivity of Dielectric Material = 36
εo = Permittivity of Vacuum = 9

εr = ε / εo
εr = 36 / 9
εr = 4

Therefore, the dielectric constant is 4.

Calculating the Permittivity of Dielectric Material when the Dielectric Constant and the Permittivity of Vacuum is Given.

ε = εro)

Where:

ε = Permittivity of Dielectric Material
εr = Dielectric Constant
εo = Permittivity of Vacuum

Let’s solve an example;
Find the permittivity of dielectric material when the dielectric constant is 14 and the permittivity of vacuum is 17.

This implies that;

εr = Dielectric Constant = 14
εo = Permittivity of Vacuum = 17

ε = εro)
ε = 14 (17)
ε = 238

Therefore, the permittivity of dielectric material is 238.