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To define Coulomb's Law or Coulomb's inverse-square law, it is an experimental law of Physics. It quantifies the amount of force between two stationary and electrically charged particles. The electric force present between the charged bodies at rest is conventionally referred to as a Coulomb force or electrostatic force. The quantity of the electrostatic force between the stationary charges is always described by Coulomb's Law.

It’s necessary for the development of electromagnetism theory. It may be a starting point because nowadays it is possible to discuss the quantity of electric charge in a meaningful way.

In 1785, for the first time, a French physicist named Charles Augustin de Coulomb coined a tangible relationship in the mathematical form between the two bodies, which are electrically been charged. He also published an equation for the force causing the bodies either to attract or repel each other, which is called Coulomb’s Law or Coulomb’s inverse-square Law.

Coulomb’s law formula can be defined both in terms of scalar and vector form.

### Scalar Form

In its scalar form, the Coulomb’s Law is given by,

F = ke \[\frac{q_{1}q_{2}}{r^{2}}\]

Where ‘ke’ is the constant of the Coulomb's Law (ke ≈ 8.99 × 109 N⋅m2⋅C−2), q1, q2 are the charges’ signed magnitudes, and the scalar ‘r’ is the distance between the charges. The interaction force between the charges is attractive if the charges have opposite signs (means F is negative) and repulsive if like-signed (means F is positive).

### Vector Form

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Considering the above representation, the Coulomb’s Law in vector form can be given by,

\[\overrightarrow{F_{12}}\] = \[\frac{1}{4 \pi \epsilon _{0}}\] - \[\frac{q_{1}q_{2}}{r_{12}^{2}}\] \[\widehat{r_{12}}\] ; \[\overrightarrow{F_{12}}\] = - \[\overrightarrow{F_{21}}\]

Here, F12 is the exerted force by q1 on q2, whereas, F21 is the force exerted by q2 on q1.

The Coulomb's Law holds for stationary charges, which are only point sized. This Law also obeys Newton’s Third Law.

(i.e \[\overrightarrow{F_{12}}\] = - \[\overrightarrow{F_{21}}\])

Force on a charged particle because of several point charges is the resultant of forces due to the individual point charges, i.e.,

\[\overrightarrow{F}\] = \[\overrightarrow{F_{1}}\] + \[\overrightarrow{F_{2}}\] + \[\overrightarrow{F_{3}}\] + …….

If ‘q’ is displaced slightly towards A, FA increases in magnitude while FB decreases in magnitude. Now the net force on ‘q’ is towards A, and so it will not return to its original position. Therefore, the equilibrium is unstable for axial displacement.

If ‘q’ is displaced perpendicular to AB, then the force FA and FB bring the charge to its original position. So for a perpendicular displacement, the equilibrium is stable.

A coulomb is a charge, which repels an equal charge of the same sign having a force of 9×109 N during the charges are apart one meter in a vacuum. Coulomb force is the conservative internal and mutual force.

The value of εo is given by,

8.86 × 10-12 C2/Nm2 or else, as, 8.86 × 10-12 Fm–1

It is to note that the Coulomb force remains true only for the static charges.

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Kr2 = constant or K1r12 = K2r22

When the force between two charges in two various media is the same for different separations, then, F = \[\frac{1}{K}\] \[\frac{1}{4 \pi \epsilon _{0}}\] \[\frac{q_{1}q_{2}}{r^{2}}\] = constant

If the force between two charges that are separated by a distance ‘r0’ in a vacuum is similar to the force between the same charges separated by a distance of ‘r’ in a medium, then based on the Coulomb’s Law; Kr2 = r02

Two spherical conductors having the charges as q1 and q2 and radii as r1 and r2 are kept in contact and thereafter separated the conductors’ charges after contact is given by

q1 = [ \[\frac{r_{1}}{r_{1}+r_{2}}\] ] (q1 + q2), q2 = [ \[\frac{r_{2}}{r_{1}+r_{2}}\] ] (q1 + q2)

Two identical conductors having q1 and q2 charges are kept into contact and then separated after each will have the same charge that is equal to \[\frac{q_{1}+q_{2}}{2}\]. If the charges are q1, -q2, then each contains a charge which is equal to \[\frac{q_{1} - q_{2}}{2}\]

If the charges are q1, -q2 then, F = F(q1 + q2)2 / 4q1q2

If the force of repulsion or attraction between two identical conductors with the charges q1, q2, when separated with a distance ‘d’, is F. If they are also kept into contact and separated by the same distance, then the new force between them is given by, F = \[\frac{F(q_{1}+q_{2})^{2}}{4q_{1}q_{2}}\]

Between the two-electrons separated by a certain distance, the Gravitational force/Electrical force = 1042

Between an electron and a proton separated by a certain distance, the Gravitational force/Electrical force = 1039

Between the two protons separated by a certain distance, the Gravitational force/Electrical force = 1036

The relationship between the permeability of free space, the velocity of light, and the permittivity of free space is described by the expression c = 1 / √ (μoεo)

Coulomb’s Law is applicable only for the point charges which are at rest

This Law can only be applied in the cases where the inverse square law is obeyed

It is difficult to implement this Law where the charges are in arbitrary shape because in those cases we cannot determine the distance between the charges

The Law cannot be used directly to calculate the charge on the big planets

FAQ (Frequently Asked Questions)

1. Explain the Applications of Coulomb’s Law?

Coulomb's Law’s applications are used to:

Calculate the force and distance between the two charges

Calculate the electric field using the formula given below as

E = F/Q_{r} (N/C)

Where 'E' is the electric field's strength, F is the Electrostatic Force, and QT is the Test charge in coulombs.

To calculate the force on one point because of the presence of several points (as a theorem of superposition)

When the Two-Point Charges, q_{1} = +9 μC, and q_{2} = 4 μC, are Separated with a Distance r = 12 cm, Calculate the Magnitude of the Electric Force.

[Given that, k = 8.988 x 10^{9} Nm^{2}C^{−2}, q_{1} = 9 ×10^{-6} C, q_{2} = 4 ×10^{-6} C, and r = 12cm = 0.12 m.]

F_{e} = k (q_{1}q_{2})/r^{2}

= (8.99 x 10^{9}(9 x 10^{-6}) (4 x 10^{-6}))/(0.12^{2})

= (8.99 x 10^{9}(3.6 x 10^{-11}))/(0.0144)

= (0.32364)/(0.01444)

Finally, F_{e} = 22.475 N.