Coulombs Law Formula

This article discusses the Coulombs law formula and its derivation. Scientists in the 18th century knew that a particle that is electrically charged would exert a certain force on another charged particle. Despite understanding this, the scientists back then had no idea about the strength of the charge. Furthermore, the scientists were clueless about the various factors affecting the strength of the charge. This problem was finally solved by Charles Coulomb when he proposed the famous Coulomb’s Law.

What is Coulomb’s Law

Coulomb’s Law provides one of the basic ideas about electricity in physics. This law takes a look at the forces which are created between two charged objects. As the distance increases then consequently there is a decrease in the forces and electric fields. The conversion of this simple idea took place into a relatively simple formula.

Coulomb’s Law is certainly an experimental law. This law quantifies the amount of force which exists between two stationary electrically charged particles. Moreover, electrostatic force or Coulomb force refers to the electric force which exists between charged bodies which are at rest. Above all, Coulomb’s Law describes the quantity of electrostatic force which is between the stationary charges.

Coulomb’s Law Formula

Coulomb’s Law finds out the magnitude of the electrostatic force between the charges. The unit of the electrostatic force is Newton (N).

Electrostatic force = (Coulomb constant) absolute value of (charge 1) (charge 2)/ (distance between charges)2

F = \(k\frac{\left | q_{1}q_{2} \right |}{r^{2}}\)
F = electrostatic force which exists between two point charges (N= kg.m/s2)
K = Coulomb constant k = \(\frac{1}{4}\pi\epsilon _{0} \cong 8.988 \times 10^{9} N.m^{2}/c^{2}\)
q1 = charge of the first point charge(C)
q2 = charge of the second point charge(C)
r = refers to the distance between the charges (m)

Coulomb’s Law Formula Derivation

Coulomb’s Law states that the separation of the two point charge q1 and q2 is by the distance ‘r’.

F happens to be directly proportional to the product of charges between them
\(F \propto q1 \times q2\)

Furthermore, F happens to be inversely proportional to the square of the distance between them
\(F \propto \frac{1}{r^{2}}\)

Together, one can arrive at
\(F \propto \frac{q1 \times q2}{r^{2}}\)
F = \(K\frac{q1q2}{r^{2}}\)

Solved Examples on Coulombs Law Formula

Q1 There are two small charged spheres which are placed 0.300 m apart. The first contains a charge of -3.00 µC (micro-Coulombs). The second contains a charge of -12.0 µC. Find out whether these charged spheres attract or repel? Also find out the magnitude of the electrostatic force on each sphere?

A1 The spheres certainly charges with the same sign. Therefore, the force between them is repulsive. Furthermore, the direction of the force on each sphere certainly points away from the other. In order to find the magnitude of the force, one must convert the charge on the particles to Coulombs. The prefix “µ”, meaning “micro”, gives an indication that the number is scaled by 10^-6, and so 1 µC = 10^-6 C. The charge of the first sphere would be:

q1 = \(-3\mu C\)
q1= \(-3\mu C\)\(\left ( \frac{10^{-6}C}{1\mu C} \right )\)
q1= -3 ×10-6 C

The charge of the second sphere is:

so, q2 = \(-12\mu C\)
q2 = \(-12\mu C\)\(\left ( \frac{10^{-6}C}{1\mu C} \right )\)
q2 = -1.2 × 10-5 C

One can find the magnitude of the electrostatic force on each sphere by making use of the Coulomb’s Law:

F = \(k\frac{\left | q_{1}q_{2} \right |}{r^{2}}\)
F = (8.988 × 10⁹N.m²/c²) \(\frac{\left ( -3 \times 10^{-6}C \right )\left ( -1.2\times 10^{-5}C \right )}{\left ( 300m \right )^{2}}\)
F = (8.988 × 10⁹N) (0.4 × 10^-9)
F \(\cong\) 3.595 × 109^-9N
F \(\cong\) 3.595 × 100N
F \(\cong\) 3.595N

Hence the magnitude of the force on each sphere happens to be 3.595N.