Coulomb's Law (original) (raw)

Last Updated : 27 May, 2026

Coulomb’s Law is defined as a mathematical concept that defines the electric force between charged objects. Coulomb's Law states that the force between any two charged particles is directly proportional to the product of the charges but is inversely proportional to the square of the distance between them. It acts along the line that connects the two charges that are regarded as point charges.

Coulomb's-Law-(1)

Coulomb's law is a mathematical formula that describes the force between two point charges. When the size of charged bodies is substantially smaller than the separation between them, then the size is not considered or can be ignored. The charged bodies can be considered point charges.

Coulomb’s Law in Scalar Form

As we know, the force (F) between two point charges q1 and q2 separated by a distance r in a vacuum is,

Proportional to the product of the charges, F ∝ q1q2

Inversely Proportional to the square of the distance between them, F ∝ 1/r2

F ∝ q1q2 / r2

then,

F = \frac { k q_1q_2} {r^2}

where,

**Coulomb’s Law in Vector Form

Coulomb's law is better written in vector notation because force is a vector quantity. Charges q1 and q2 have location vectors r1 and r2, respectively. F12 denotes the force on q1 owing to q2, and F21 denotes the force on q2 owing to q1. For convenience, the two-point charges q1 and q2 have been numbered 1 and 2, respectively, and the vector leading from 1 to 2 has been designated by r21.

Coulomb’s Law in Vector Form

\overrightarrow{r}_{21} = \overrightarrow{r}_2– \overrightarrow{r}_1

Similarly, the vector leading from 2 to 1 is denoted by r12,

\overrightarrow{r}_{12} = \overrightarrow{r}_1– \overrightarrow{r}_2

r21 and r12 are the magnitudes of the vectors and \overrightarrow{r}_{12}, respectively and magnitude r12 is equal to r21. A unit vector along the vector specifies the vector's direction. The unit vectors are used to denote the direction from 1 to 2 (or 2 to 1). The unit vectors are defined as,

\hat{r}_{21}=\dfrac{\overrightarrow{{r}}_{21}}{r_{21}}

Similarly,

\hat{r}_{12}=\dfrac{\overrightarrow{{r}}_{12}}{r_{12}}

Coulomb’s force law between two point charges q1 and q2 located at vectors r1 and r2 is then expressed as,

\begin{aligned}\overrightarrow{F}_{21}&=\dfrac{1}{4\pi{\epsilon}_\circ}\dfrac{q_1q_2}{{r}_{21}^2}\hat{r}_{21}\\&=\frac{1}{4\pi{\epsilon}_\circ}\frac{q_1q_2}{{r}_{21}^3}\overrightarrow{r}_{21}\end{aligned}

**Key Points on Coulomb’s Law

\begin{aligned}\overrightarrow{F}_{21}&=\dfrac{1}{4\pi{\epsilon}_\circ}\dfrac{q_1q_2}{{r}_{21}^2}\hat{r}_{21}\\ &=\frac{1}{4\pi{\epsilon}_\circ}\frac{q_1q_2}{{r}_{21}^3}\overrightarrow{r}_{21}\end{aligned}

Conditions for Stability of Coulomb’s Law

If two charges are arranged in a straight line AB, and one charge q is slightly displaced towards A, the force acting on A FA increases in magnitude while the force acting on B FB decreases in magnitude. Thus, the net force on q shifts towards A. So we can say that for axial displacement, the equilibrium is unstable.

If q is displaced perpendicular to line AB, the forces FA and FB are changed in such a manner that they bring the charge to its original position. Now we can say that for perpendicular displacement, the equilibrium is stable.

Applications of Coulomb’s Law

Coulomb’s Law is one of the basic laws of Physics. It is used for various purposes; some of its important applications are discussed below.

An electric field is given by,

**E = F / Q T (N/C)

where,

Limitations of Coulomb’s Law

There are some limitations of Coulomb’s Law, which are discussed below:

Solved Problems

**Example 1: Charges of magnitude 100 micro coulombs each are located in a vacuum at the corners A, B, and C of an equilateral triangle measuring 4 meters on each side. If the charges at A and C are positive and the charge at B is negative, what is the magnitude and direction of the total force on the charge at C?

**Solution:Example 1 - Columbs Law

Force FCA is applied toward AC, and the expression for the FCA is expressed as

F_{CA}=\dfrac{qq}{4\pi{\epsilon}_\circ}

Substitute the values in the above expression,

F_{CA}=\dfrac{100\times10^{-6}\times100\times10^{-6}}{4\pi\times8.854\times10^{-12}}\\ F_{CA}=5.625\text{ N}

The Force FCB is applied toward CB, and the expression for the FCB is expressed as

F_{CB}=\dfrac{qq}{4\pi{\epsilon}_\circ}

Substitute the values in the above expression,

F_{CB}=\dfrac{100\times10^{-6}\times100\times10^{-6}}{4\pi\times8.854\times10^{-12}}\\ F_{CB}=5.625\text{ N}

Therefore, the two forces are equal in magnitude but in different directions. The angle between them is 120º. The resultant force F is given by,

F=\sqrt{F_{CA}^2+F_{CB}^2+2F_{CA}F_{CB}\cos\theta}\\ F=\sqrt{5.625^2+5.625^2+2\times5.625\times5.625\times\cos120^\circ}\\ F=5.625\text{ N}

**Example 2: A positive charge of 6×10-6 C is 0.040 m from the second positive charge of 4×10-6 C. Calculate the force between the charges.

**Solution: Given,

First charge q1 = 6×10-6 C.

Second charge q2 = 4×10-6 C.

Distance between the charges r = 0.040 m

k = 9×109

We know that, **F = k q 1 q 2 ****/ r** 2

Substitute the values in the above expression,

F = k q1q2 / r2

F = 9×109×[(6×10-6)× (4×10-6)] / (0.04)2

F= 134.85 N

**Example 3: Two-point charges, q1 = +9 μC and q2 = 4 μC, are separated by a distance r = 12 cm. What is the magnitude of the electric force?

**Solution: Given,

**F = k (q 1 q 2 **∕ r 2 )

F = (8.9875 × 109 ) [(9x 10-6 ) × (4 x 10-6) / (0.12)2]

F = (8.9875 × 109 ) [36 × 10-12 /0.0144]

**F = 22470 N

The electric force between the charges is approximately **22.47 N