Table of contents 
What is Coulomb’s Law? 
Key Points on Coulomb’s Law 
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According to Coulomb’s law, the force of attraction or repulsion between two charged bodies is directly proportional to the product of their charges and inversely proportional to the square of the distance between them. It acts along the line joining the two charges considered to be point charges.
Coulomb's Law Formula
A French physicist Charles Augustin de Coulomb in 1785 coined a tangible relationship in mathematical form between two bodies that have been electrically charged. He published an equation for the force causing the bodies to attract or repel each other which is known as Coulomb’s law or Coulomb’s inversesquare law.
Coulomb’s law holds for stationary charges only which are point sized. This law obeys Newton’s third law
Force on a charged particle due to a number of point charges is the resultant of forces due to individual point charges i.e.
What is 1 Coulomb of Charge? A coulomb is that charge which repels an equal charge of the same sign with a force of 9 × 10^{9} N, when the charges are one meter apart in a vacuum. Coulomb force is the conservative mutual and internal force.
The value of ε_{o} is 8.86 × 10^{12} C^{2}/Nm^{2} (or) 8.86 × 10^{12} Fm^{–1}Note: Coulomb force is true only for static charges.
If q is slightly displaced towards A, F_{A} increases in magnitude while F_{B} decreases in magnitude. Now the net force on q is toward A so it will not return to its original position. So for axial displacement, the equilibrium is unstable.
If q is displaced perpendicular to AB, the force F_{A} and F_{B} bring the charge to its original position. So for perpendicular displacement, the equilibrium is stable.
The force between 2 charges depends upon the nature of the intervening medium, whereas gravitational force is independent of the intervening medium.
For air or vacuum, since for air or vacuum, ε_{r} = K = 1
The value of is equal to 9 × 10^{9} Nm^{2}/C^{2}.
Where E = Strength of the electric field
F = Electrostatic force
Q_{T} = Test charge in coulombs
1. If the force between two charges in two different media is the same for different separations, = constant.
2. Kr^{2} = constant or K_{1}r_{1}^{2} = K_{2}r_{2}^{2 }
3. If the force between two charges separated by a distance ‘r_{0}’ in a vacuum is the same as the force between the same charges separated by a distance ‘r’ in a medium, then from Coulomb’s Law; Kr^{2} = r_{0}^{2}
4. Two identical conductors having charges q_{1} and q_{2} are put to contact and then separated after which each will have a charge equal to q_{1} + q_{2}/2. If the charges are q_{1} and – q_{2}, then each will have a charge equal to q_{1}  q_{2}/2.
5. Two spherical conductors having charges q_{1} and q_{2} and radii r_{1} and r_{2} are put to contact and then separated the charges of the conductors after contact is;
q_{1} = [r_{1}/(r_{1} + r_{2})] (q_{1} + q_{2}) and q_{2} = [r_{2}/(r_{1} + r_{2})] (q_{1} + q_{2})
6. If the force of attraction or repulsion between two identical conductors having charges q_{1} and q_{2} when separated by a distance d is F. Also if they are put to contact and then separated by the same distance the new force between them is
7. If charges are q_{1} and q_{2} then, F = F(q_{1} + q_{2})^{2} / 4q_{1}q_{2}
8. Between two electrons separated by a certain distance: Electrical force/Gravitational force = 10^{42}
9. Between two protons separated by a certain distance: Electrical force/Gravitational force = 10^{36}
10. Between a proton and an electron separated by a certain distance: Electrical force/Gravitational force = 10^{39}
11. The relationship between the velocity of light, the permeability of free space and permittivity of free space is given by the expression c = 1 / √ (μ_{o}ε_{o})
12. If Coulomb’s law is applied to two identical balls of mass m are hung by silk thread of length ‘l’ from the same hook and carry similar charges q then;
Example 1: Charges of magnitude 100 microcoulomb each are located in vacuum at the corners A, B and C of an equilateral triangle measuring 4 meters on each side. If the charge at A and C are positive and the charge B negative, what is the magnitude and direction of the total force on the charge at C?
The situation is shown in fig. Let us consider the forces acting on C due to A and B.
Now, from Coulomb’s law, the force of repulsion on C due to A i.e., FCA in direction AC is given by
The force of attraction on C due to B i.e., F_{CB} in direction CB is given by
Thus the two forces are equal in magnitude. The angle between them is 120º. The resultant force F is given by
This force is parallel to AB.
Example 2: The negative point charges of unit magnitude and a positive point charge q are placed along the straight line. At what position and for what value of q will the system be in equilibrium? Check whether it is stable, unstable or neutral equilibrium.
The two negative charges A and B of unit magnitude are shown in fig. Let the positive charge q be at a distance r_{A} from A and at a distance r_{B} from B.
Now, from coulombs law, Force on q due to AForce on q due to B
These two forces acting on q are opposite and collinear. For the equilibrium of q, the two forces must also be equal i.e.
F_{q}A = F_{q}B
or
Hence rA = rB
So for the equilibrium of q, it must be equidistant from A & B i.e. at the middle of AB Now for the equilibrium of the system, A and B must be in equilibrium. For the equilibrium of A
Force on A by q = towards q
Force on A by B =
away from q
The two forces are opposite and collinear. For equilibrium the forces must be equal, opposite and collinear. Hence
or q = 1/4 in magnitude of either charge.
It can also be shown that for the equilibrium of B, the magnitude of q must be 1/4 of the magnitude of either charge.
Example 3: A positive charge of 6 × 10^{6} C is 0.040m from the second positive charge of 4 × 10^{6} C. Calculate the force between the charges.
Given
q_{1} = 6 × 10^{6} C
q_{2} = 4 × 10^{6} C
r = 0.040 m
F_{e} = 134.85 N
Example 4: Twopoint charges, q_{1} = +9 μC and q_{2} = 4 μC, are separated by a distance r = 12 cm. What is the magnitude of the electric force?
given
k = 8.988 x 10^{9} Nm^{2}C^{−2}
q1 = 9 ×106 C
q2 = 4 ×106 C
F_{e} = 22.475 N
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