Coulomb’s Law , Force between two point charges

Coulomb’s Law – Force between two point electric charges:

The electrostatic force of interaction (attraction or repulsion) between two point electric charges is directly proportional to the product of the charges , inversely proportional to the square of the distance between them and acts along the line joining the two charges.

The force acting between two point charges (q1 and q2) separated by a distance r is given as

F ∝ q1q2

$ \displaystyle F\propto \frac{1}{r^2} $

$ \displaystyle F\propto \frac{q_1 q_2}{r^2} $

$ \displaystyle F = k \frac{q_1 q_2}{r^2}$

$ \displaystyle k = \frac{1}{4\pi\epsilon_0} \: $ where εo is the electrical permittivity of free space in which the two charges are kept.

Also ε = εo εr = Absolute Permittivity of the medium.

Where εo = 8.85 × 10-12 C2N-1m-2 is the permittivity of the free space

and εr = relative permittivity of the medium.

For vacuum (free space) εr = 1 and

$ \displaystyle k = \frac{1}{4\pi \epsilon_0} = 9 \times 10^9 N m^2 C^{-2}$

Coulomb’s Law is basically the force (F ) exerted by one electric charge on the other independent of the presence or absence of a dielectric medium containing the charge.

The net force on a charge is the superposition of Coulomb’s force and the force produced by the polarized dielectric medium.

The net force is

$ \displaystyle F = \frac{1}{4\pi\epsilon} \frac{q_1 q_2}{r^2}$

$ \displaystyle F = \frac{1}{4\pi\epsilon_0\epsilon_r} \frac{q_1 q_2}{r^2}$

Relative Permittivity or Dielectric Constant or Specific Inductive Capacity or Dielectric Coefficient:

The dielectric constant or relative permittivity or specific inductive capacity or dielectric coefficient is given by the ratio of the absolute permittivity of the medium to the permittivity of free space.

$ \displaystyle K = \epsilon_r = \frac{\epsilon}{\epsilon_0}$

The dielectric constant or relative permittivity or specific inductive capacity or dielectric coefficient can also be defined as the ratio of the electrostatic force between two charges separated by a certain distance in vacuum to the electrostatic force between the same two charges separated by the same distance in that medium.

$ \displaystyle K = \epsilon_r = \frac{F_0}{F_m}$

Dielectric constant has no unit.

Similarity with Newton’s Law of Gravitational Attraction:

1. Both obey inverse square law.

2. Both follow principle of superposition.

3. The forces are central force, i.e. they act along the line joining the particles.

Dissimilarity with Newton’s Law of Gravitational Attraction:

1. The net force on an electric charge is dependent on the medium , whereas the Newton’s force of gravitational attraction is independent of the medium.

2. Coulomb force may be attractive or repulsive but Newton’s Gravitational force is always attractive.

Example : Consider three charges q1, q2 and q3 each equal to q at the vertices of an equilateral triangle of side ‘l’ what is the force on any charge due to remaining charges.


Force acting on C due to A is

$\large F_1 = \frac{1}{4\pi \epsilon_0}\frac{q^2}{l^2}$

Force acting on C due to B is

$\large F_2 = \frac{1}{4\pi \epsilon_0}\frac{q^2}{l^2}$

Net force acting on a charged particle at C is

$\large F = \sqrt{F_1^2 + F_2^2 + 2 F_1 F_2 cos60^o}$

$\large F = \sqrt{F_1^2 + F_1^2 + 2 F_1^2 (1/2)}$ (since F1 = F2)

$\large F = \sqrt{3 F_1^2} = \sqrt{3} F_1$

$\large = \sqrt{3}\frac{1}{4\pi \epsilon_0}\frac{q^2}{l^2}$

Example : Two particles A and B having charges 8 x 10-6C and – 2 x10-6C respectively are held fixed with a separation of 20 cm. Where should a third charged particle C be placed so that it does not experience a net electric force ?

Solution : As the net electric force on C should be equal to zero, the force due to A and B must be opposite in direction. Hence, the particle should be placed on the line AB.

As A and B have charges of opposite signs, C cannot be between A and B Also A has larger magnitude of charge than B. Hence, C should be placed closer to B than A.

Suppose BC = x and the charge on C is Q

$ \displaystyle \vec{F_{CA}} = \frac{1}{4\pi\epsilon_0 } \frac{(8\times10^-6)Q \hat{i}}{(0.2+x)^2 }$

and , $ \displaystyle \vec{F_{CB}} = -\frac{1}{4\pi\epsilon_0 } \frac{(2\times10^-6)Q \hat{i}}{(x)^2 }$

$ \displaystyle \vec{F_C} = \vec{F_{CA}} + \vec{F_{CB}} $

But Fc = 0

Hence , $ \displaystyle \vec{F_{CA}} = \frac{1}{4\pi\epsilon_0 } [\frac{(8\times10^-6)Q \hat{i}}{(0.2+x)^2 } – \frac{(2\times10^-6)Q \hat{i}}{(x)^2 }] = 0$

Which gives x = 0.2 m

Illustration : A particle of mass ‘m’ carrying a charge -q1 is moving around a fixed charge +q2 along a circular path of radius ‘r’ find time period of revolution of charge q

Sol:  Electrostatic force on -q1 to +q2 will provide the necessary centripetal force

$\large \frac{1}{4\pi \epsilon_0} \frac{q_1 q_2}{r^2} = \frac{m v^2}{r}$

$\large v = \sqrt{\frac{q_1 q_2}{4\pi \epsilon_0 m r}}$

$\large T = \frac{2 \pi r}{v}$

$\large T = \sqrt{\frac{16 \pi^3 \epsilon_0 m r^3}{q_1 q_2}}$

Also Read :

→ Charging of the body & Properties of Charge
→ Principle of superpostion
→ Electric field
→ Electric Potential Energy
→ Electric Lines of Force & its Properties
→ Electric Potential
→ Equipotential Surface , Relation b/w Electric field & Potential
→ Electric Dipole , Dipole Moment & Electric Field due to Dipole
→ Electric Flux , Gauss’ Law & Applications of Gauss’s Law
→ Charge appearing on the Surface of Plates
→ Mechanical Force & Electric Pressure on a Charged Surface

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