“The force of attraction or repulsion between two charges varies to the product of the two charges and inversely proportional to the square distance between them.” Coulomb’s law states.
In 1785, Charles-Augustin de Coulomb(A french physicist) proposed the quantitative theory of the two-point charges in electricity. He termed point charge as When a linear size of charged bodies is much smaller than the distance between the charge bodies is called point charge.
Coulomb’s law states,
“The electrostatic force between two point charges is directly proportional to the product of charges and varies inversely to the distance square.”
To understand this chapter, students can refer to NCERT Solution for class 12 physics.
Coulomb’s Law Formula
As per coulomb’s law, the force of attraction between two charges varies directly to the product of charges. It is inversely proportional to the square of the distance.
Where
F = Electrostatic force
K = Coulomb’s constant and its value is 1/4 π ε0
q1&q2 = Point charges
r = Distance between the two charges
Coulomb’s law Derivation
Consider the point charges, q1&q2 and they are separated by the distance r.
Then according to coulomb’s law, the force of attraction is directly proportional to the product of the charges.
F ∝ q1q2
And is inversely proportional to the distance’s square.
F ∝ 1r2
Then
F ∝ q1q2 r2
| F = Kq1q2 r2 |
Where K is Coulomb’s constant and its value is equal to the 1/4 π ε0.
Vector Representation of Coulomb’s Law
We can represent the coulomb’s law in vector form here because force is a vector quantity with magnitude and direction.
Consider the two charges q1&q2 and r1&r2 are the position vector of the given charge.
The same polarity repels, and the opposite attracts. The force on F12 is due to force of q2 on, q1 and the force on F21 is due to force of q1 on q2.
Then, the vector representation from q1 to q2 is, r21 i.e. r2 – r1.
So the vector representation of the vector from r1 to r2 and r2 to r1 is
The force act on q2 due to q1 in the vector form will be
The above equation is the vector representation of coulomb’s law.
- Here, the force exerted on F12 is due to force of q2 on, q1 and the force on F21 is due to force of q1 on q2.
- Coulomb’s law follows newton’s third law.
F12 = –F21
- The resultant force due to the number of individual charge particles is
F = F1 + F2 + F3 + F4………
Limitations of Coulomb’s law
- This law only applies in cases where the inverse square law is followed.
- This law is only applicable in the case of point charges at rest.
- Finding the force of attraction between the charges in arbitrary shapes is difficult.
- We can’t find the charge on different planets with the help of the coulomb’s law.
Application of coulomb’s law
- It is used to find the force of attraction between two-point charges.
- It is used to calculate the electric field.
- The superposition theorem is used to find the force on one point charge due to several other charges.
Examples of Coulomb’s Law
Question1. Two small charged spheres having charges of 2 × 10−7 C and 3 × 10−7 C are placed 30 cm apart in the air. What is the force between them?
Solution: The formula used to find the force, F is given as,
Where, q1 and q2 are the charges.
r is the distance between the charges.
is a constant and its value is
is the permittivity of free space. Its value is
Now, the diagram is:
Since, both the charges are positive, thus, the nature of force will be repulsive. F12 the force on charge q1 caused by charge q2.
Now, Given:
Charge on the first sphere,
Charge on the second sphere,
Distance between the spheres, r = 30 cm = 0.3 m
Putting the values in equation (1), we get,
Hence, the force between the given charged particles will be
Since the nature of the charges is the same i.e. they are both positive. Hence, the force will be repulsive.
Question2. In the air, a 0.2 N electrostatic force on a small sphere of charge of 0.4 μC due to another small sphere of charge of 0.8 μC.
(a) Find the distance between the two spheres?
(b) Find the force on the second sphere due to the first?
Solution: (a) Electrostatic force on the first sphere, F = 0.2 N
Charge on this sphere, q1 = 0.4 μC = 0.4 × 10−6 C
Charge on the second sphere, q2 = − 0.8 μC = − 0.8 × 10−6 C
Electrostatic force between the spheres is given by the relation,
Since where, ∈0 = Permittivity of free space
The distance between the two spheres is 0.12 m.
(b) Both the spheres attract each other with the same force. Therefore, the force on the second sphere due to the first is 0.2 N.
Question3. A positive charge of 6×10-6 C is placed 0.040m is the second positive charge of 4×10-6 C. Find the force between the two charges.
Solution q1 = 6×10-6 C
q2 = 4×10-6 C
r = 0.040 m
Frequently Asked Questions(FAQs)
What is coulomb’s law?
According to the coulomb’s law, the electrostatic force between the two point charges is directly proportional to the product of two charges and inversely proportional to the square of the distance between the charges.
Here,
F ∝ q1q2 r2
F = Kq1q2 r2
What are the limitations of coulomb’s law?
Here are the limitations of coulomb’s law
- It is only applicable in point charges at rest.
What is the value of a one-coulomb charge?
The value of a one-coulomb charge is 9 109.
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