Coulomb’s Law
What is Coulomb’s Law
All charged particles attract or repel one another depending upon the nature of charges., i.e., like charges repel and unlike charges attract. Coulomb’s law governs the electrostatic force of attraction or repulsion between them. The law is usually applied to point charges. It gives a relationship between the electrostatic force, the magnitude of the charges, and the separation distance.
Statement of Coulomb’s Law
Coulomb’s law states that the electrostatic force between the two particles is,
 Directly proportional to the product of the magnitude of the charges
 Inversely proportional to the square of the distance betweenthe two charges
This law is named after French physicist CharlesAugustin de Coulomb, who published it in 1785.
Coulomb’s Law Equation
Scalar Form
Suppose two point charges are separated by a certain distance. Then, according to Coulomb’s law, the electrostatic force, also known as Coulomb’s force, between the two charges is given by the following equation.
Formula: F = k q_{1} q_{2}/r^{2}
Where,
F: Electrostatic force
k: Coulomb’s constant
q_{1}, q_{2}: Magnitude of the point charges
r: Distance of separation between the charges
SI Unit: Newton or N
Dimension: [M L T^{2}]
Because of the inverse relationship, this law is sometimes called Coulomb’s inverse square law.
Vector Form
The vector form of Coulomb’s law is given by,
F = k q_{1} q_{2}/r r
Where,
r: Unit vector along a line connecting the two charges
Coulomb’s Law Constant
The constant in Coulomb’s law is known as Coulomb’s constant, which is denoted by k. Suppose, q_{1} = 1, q_{2} = 1, and r = 1, then,
F = k 1 x 1/1 = k
Therefore, the Coulomb’s constant is defined as the electrostatic force experienced by two unit charges when a unit distance separates them. It is expressed as follows.
In free space,
k = 1/4πε_{o}
In a medium,
k = 1/4πε_{r}
Where,
ε_{o}: Permittivity of free space or vacuum (= 8.85 x 10^{12} C^{2}N^{1}m^{2})
ε_{r}: Permittivity of the medium
The ratio of ε_{r} and ε_{o} is known as the relative permittivity of the medium or dielectric constant.
Dielectric Constant: κ = ε_{r}/ε_{o}
Putting the value of ε_{o} gives the value of k,
Coulomb’s Constant: 9 x 10^{9} Nm^{2}C^{2}
From the expression for k, the Coulomb’s Law equation in free space is given by,
F = (1/4πε_{o}) q_{1} q_{2}/r^{2}
Superposition Principle of Coulomb’s Law
Coulomb’s law refers to the interaction between two point charges. For more than two charges, the force on each charge due to the presence of other charges should be taken into account. Since force is a vector quantity, the superposition principle must be applied.
According to the superposition of forces, all linear forces can be added vectorially. The net force is the vectorial sum of all the individual forces. Consider a system of n charges given by q_{1}, q_{2}, …, q_{n}, the force due to charges q_{1} and q_{2} is,
F_{12} = k q_{1} q_{2}/r_{12} r_{12}
Where,
r_{12}: distance of separation between the two charges
r_{12}: unit vector along the direction of the line connecting the two charges
Similarly, the force between q_{1} and any other charge is,
F_{13}, F_{14}, …, F_{1,n}
Therefore, the net force due to q_{1} is,
F_{1} = F_{12} + F_{13} + … F_{1,n}
Limitations
 Applicable only to point charges at rest
 Applicable to cases where inversesquare law is obeyed
 It cannot be applied to irregularshaped charges
 It cannot be applied to a continuous charge distribution
Applications
 To calculate the force between two charges
 To calculate the distance between two charges
 To study the motion of electrons around an orbit in an atom
 To calculate the electrostatic potential and hence the work done in moving a charge from one point to another
 To calculate the relative permittivity (dielectric constant) of a medium
 Inkjet printers use electrostatic forces to spray tiny ink droplets, which are charged when passing through a nozzle
 Photocopiers and laser printers use an electrostatic process in which an electrically charged drum coated with selenium attracts a blank piece of paper to print images
Other applications include power coating and electrostatic air cleaning
Examples and Practice Problems
P.1. Two identical charges 6.25 nC are held apart at a separation distance of 42 cm. Determine the magnitude of the electrical force of repulsion between them.
Soln.: Given,
q_{1} = 6.25 nC
q_{2} = 6.25 nC
r = 42 cm = 0.42m
We have,
F = k q_{1} q_{2} /r^{2}
= (9 x 10^{9} Nm^{2}C^{2} x 6.25 nC x 6.25 nC)/0.42^{2} m^{2}
= 2 μN
P.2. The force between two identical charges separated by 2 cm is equal to 45 N. What is the magnitude of the two charges?
Soln.: Given,
F = 45 N
d = 2 cm = 0.02 m
q = q_{1} = q_{2}
We have,
F = k q_{1} q_{2} /r^{2}
= k q^{2}/r^{2}
Or, q = sqrt (F r^{2}/k)
= sqrt (45 N x (0.02 m)^{2}/ 9 x 10^{9} Nm^{2}C^{2})
= 1.41 μC
P.3. Two balloons with charges of +4 µC and 8 µC attract each other with a force of 0.0625 N. Determine the separation distance between the two balloons.
Soln.: Given,
F = 0.0625 N
q_{1} = 4 μC = 4 x 10^{6} C
q_{2}= 8 μC = 8 x 10^{6} C
We have,
F = k q_{1} q_{2} /r^{2}
Or, r = sqrt (k q_{1} q_{2 }/ F)
= sqrt (9 x 10^{9} Nm^{2}C^{2} x 4 x 10^{6} C x 8 x 10^{6} C/0.0625 N)
= 2.15 m
P.4. Two neutrally charged bodies are separated by 0.5 cm. Electrons are stripped off from one body and placed on the second body until a force of 2 μN is generated between them. How many electrons were transferred between the bodies?
Soln. Given,
q_{1} = +ne
q_{2} = ne
e = 1.6 x 10^{19} C
r = 0.5 cm = 5 x 10 ^{3} m
F = 2 μN = 2 x 10^{6} N
We have,
F = k q_{1} q_{2} /r^{2}
Or,F = k n^{2}e^{2} /r^{2}
Or, n = sqrt (F r^{2} / k e^{2})
= sqrt (2 x 10^{6} N x (5 x 10 ^{3} m)^{2} / (9 x 10^{9} Nm^{2}C^{2} / (1.6 x 10^{19} C)^{2}))
= 6.5 x 10^{8}

References
Article was last reviewed on Wednesday, June 16, 2021