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Coulomb's Law

History

The first measurement of the force between electric charges was made in 1784 AD by Charles Coulomb. He was a French military engineer. On the basis of these measurements, he deduced a law known as Coulomb's Law.

Introduction

We know that there are two types of charges, namely positive and charges. The charge on an electron is assumed to be negative and charge on a proton is positive. Moreover we know that like charges repel each other while opposite charges attract each other.

Definition

The force between two point charges is directly proportional to the product of the magnitudes of the charges and inversely proportional to the square of hte distance between them. It is mathematically expressed as
\[F \propto \,\frac{{{q_1}{q_2}}}{{{r^2}}}\] or \[F = k\,\frac{{{q_1}{q_2}}}{{{r^2}}}\]

Explaination

Where F is the magnitude of the mutual force that acts on each of the two point charges

\[{q_1}\]

and

\[{q_2}\]

and r is the distance between tham. The force F always acts along the line joining the two point charges, K isthe constant of proportionality. Its value depends upon the nature of medium between the two charges and system of units in which F, q and r are measured. If the medium between the two point charges is free space and the system of units is SI, then k is represented as \[k\, = \,\frac{1}{{4\pi {\varepsilon _o}}}\] where

\[{{\varepsilon _o}}\]

is an electrical constant, known as permitivitiy of free space. In SI units, its value is

\[8.85{\mkern 1mu} \times {\mkern 1mu} {10^{ - 12}}N{m^2}{C^{ - 2}}\]

. Substituting the value of

\[{{\varepsilon _o}}\]

the constant

\[k{\mkern 1mu} = {\mkern 1mu} \frac{1}{{4\pi {\varepsilon _o}}}\, = 9 \times {\mkern 1mu} {10^9}N{m^2}{C^{ - 2}}\]

Coulomb's Law in free space

Coulomb's Law in free space can be written as

\[F = {\mkern 1mu} \,\frac{1}{{4\pi {\varepsilon _o}}}\frac{{{q_1}{q_2}}}{{{r^2}}}\,\]

As stated earlier, Coulomb's force is mutual force, it means that if

\[{{q_1}}\]

exerts a force on

\[{{q_2}}\],

then

\[{{q_2}}\]

also exerts an equal and opposite force on

\[{{q_1}}\].

If we denote the force exerted on

\[{{q_2}}\]

by

\[{{q_1}}\]

as

\[{\vec F_{12}}\]

and that on charge

\[{{q_1}}\]

due to

\[{{q_2}}\]

as

\[{\vec F_{21}}\]

then

\[{{\vec F}_{12}}\, = \, - {{\vec F}_{21}}\] The magnitude of both these two forces is the same and is given by

Exercise 4.3 Question 2(i)F Sc Second Year Chapter 4 Punjab board Pakistan

Exercise 5.1 Question 8 F.Sc First Year Chapter 5 Punjab board Pakistan

Exercise 5.1 Question 7 F.Sc First Year Chapter 5 Punjab board Pakistan