How did Coulomb determine the law of electric force between two point charges?

Vedclass pdf generator app on play store
Vedclass iOS app on app store
(N/A) Coulomb assumed the charge on a metallic sphere is $q$. If the sphere is placed in contact with an identical uncharged sphere,the charge will be distributed equally over the two spheres. By symmetry,the charge on each sphere will be $\frac{q}{2}$.
By repeating this process,we can obtain charges of $\frac{q}{2}, \frac{q}{4}, \frac{q}{8}, \dots$ on the spheres.
Coulomb varied the distance $r$ for a fixed pair of charges and measured the force $F$ for different separations. He observed the relation:
$F \propto \frac{1}{r^{2}} \quad (1)$
He then varied the charges $q_{1}$ and $q_{2}$ while keeping the distance fixed. By comparing the forces for different pairs of charges,he established the relation:
$F \propto q_{1} q_{2} \quad (2)$
Combining these,the electric force between two point charges is given by:
$F \propto \frac{q_{1} q_{2}}{r^{2}}$
Therefore,$F = k \frac{q_{1} q_{2}}{r^{2}}$,where $k$ is the Coulomb constant.

Explore More

Similar Questions

$ABC$ is a right-angled triangle in which $AB = 3\,cm$ and $BC = 4\,cm$. And $\angle ABC = \pi / 2$. The three charges $+15\,e.s.u.$,$+12\,e.s.u.$,and $-20\,e.s.u.$ are placed respectively on $A$,$B$,and $C$. The force acting on $B$ is.......$dynes$.

Difficult
View Solution

Two spheres of radii $3 \ cm$ and $1 \ cm$ are placed at a distance of $10 \ cm$ from each other in a vacuum. If each sphere is charged to a potential of $10 \ V$,the force of repulsion between them is:

Positive point charges are placed at the vertices of a star shape as shown in the figure. The direction of the electrostatic force on a negative point charge at the centre $O$ of the star is

Equal charges $q$ are placed at the four corners $A, B, C, D$ of a square of side length $a$. The magnitude of the net electrostatic force on the charge at $B$ is:

Difficult
View Solution

$A$ charge $Q$ is divided into two parts $q$ and $Q - q$. If the Coulomb repulsion between them when they are separated by a distance $r$ is to be maximum,the ratio of $\frac{Q}{q}$ should be:

Difficult
View Solution

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo