** Coulomb’s Law** is defined as a mathematical concept that defines the electric force between charged objects. Columb’s Law states that the force between any two charged particles is directly proportional to the product of the charge but is inversely proportional to the square of the distance between them. Let’s learn about Columb’s law in detail in this article.

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## What is Coulomb’s Law?

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.

Force of attraction or repulsion between two charged things is directly proportional to the product of their charges and inversely proportional to the square of the distance between them, according to

. It acts along the line that connects the two charges that are regarded to be point charges.Coulomb’s law

Coulomb studied the force between two point charges and found that it is inversely proportional to the square of the distance between them, directly proportional to the product of their magnitudes, and acting in a line that connects them.

## History of Coulomb’s Law

Charles Augustin de Coulomb a French mathematician in 1785 first describes a force between two charged bodies in mathematical equations. He stated that the charge bodies repel or attract each other accordingly based on their charge, i.e. opposite charge attracts each other and similar charge repels. He also states the mathematical formula for the force between them, which is called Columb’s Law.

## Coulomb’s Law Formula (Scalar Form)

As we know, the force (F) between two point charges q_{1} and q_{2} separated by a distance r in a vacuum is,

Proportional to the product of the charges.

**F ∝ q**_{1}**q**_{2}

Inversely Proportional to the square of the distance between them,

**F ∝ 1/r**^{2}

**F ∝ q**_{1}**q**_{2}** / r**^{2}

then,

F = k q_{1}q_{2 }/ r^{2}

where,

is proportionality constant and equals to**k****1/4πε**_{0}**.**- Symbol
**ε**is_{0}.**permittivity of a vacuum** - Value of
is**k****9 × 10**^{9}**Nm**^{2}**/ C**{when we take the S.I unit of value of ε^{2}_{0}is**8.854 × 10**^{-12}**C**^{2}**N**^{-1}**m**.}^{-2}

**Coulomb’s Law in Vector Form**

**Coulomb’s Law in Vector Form**

Coulomb’s law is better written in vector notation because force is a vector quantity. Charges q_{1} and q_{2} have location vectors r_{1} and r_{2}, respectively. F_{12} denotes force on q_{1} owing to q_{2} and F_{21} denotes force on q_{2} owing to q_{1}. For convenience, the two-point charges q_{1} and q_{2} have been numbered 1 and 2, respectively, and the vector leading from 1 to 2 has been designated by r_{21}.

[Tex]\overrightarrow{r}_{21} = \overrightarrow{r}_2– \overrightarrow{r}_1 [/Tex]

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

_{12,}[Tex]\overrightarrow{r}_{12} = \overrightarrow{r}_1– \overrightarrow{r}_2 [/Tex]

r

_{21}and r_{12}are the magnitudes of the vectors [Tex]\overrightarrow{r}_{21} [/Tex] and [Tex]\overrightarrow{r}_{12} [/Tex], respectively and magnitude r_{12}is equal to r_{21}. 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 define as,[Tex]\hat{r}_{21}=\dfrac{\overrightarrow{{r}}_{21}}{r_{21}} [/Tex]

Similarly,

[Tex]\hat{r}_{12}=\dfrac{\overrightarrow{{r}}_{12}}{r_{12}} [/Tex]

Coulomb’s force law between two point charges q

_{1}and q_{2}located at vector r_{1}and r_{2}is then expressed as,[Tex]\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} [/Tex]

**Key Points on Coulomb’s Law**

**Key Points on Coulomb’s Law**

- Coulomb’s Law holds true regardless of whether q
_{1}and q_{2}are positive or negative. F_{21 }is toward [Tex]\hat{r}_{21} [/Tex], which is a repulsive force, as it should be for like charges that are if q_{1}and q_{2}are of the same sign (either both positive or both negative). When the signs of q_{1}and q_{2}are opposite or dislike charges, F_{21}is toward [Tex]-\hat{r}_{21} [/Tex], that is toward [Tex]\hat{r}_{12} [/Tex] which shows attraction, as expected for dissimilar charges. As a result, we don’t need to construct separate equations for like and unlike charges. Both instances are handled correctly by the above expression for Coulomb’s force law. - Coulomb’s force law can be used to calculate the force F
_{12}on charge q_{1}due to charge q_{2}by simply swapping 1 and 2 as,

[Tex]\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} [/Tex]

- Coulomb’s law agrees with Newton’s third law.
- In a vacuum, Coulomb’s law expression determines the force between two charges q
_{1}and q_{2}. If the charges are deposited in matter or there is matter in the intervening area, the situation becomes more complicated due to the presence of charged matter constituents. - Two identical conductors with charges q
_{1}and q_{2}are brought into contact and subsequently separated, resulting in each conductor having a charge equal to (q_{1}+q_{2})/2. Each charge will be equal to (q_{1}-q_{2})/2 if the charges are q_{1}and –q_{2}.

## What is 1 Coulomb of Charge?

Columb is the SI unit of charge. If a charge repels an equal charge of the same sign with a force of 9×10^{9} N the charge is of 1 Coulomb given so the charges are one meter apart in a vacuum.

1 coulomb is a bigger unit of charge and is not used in daily life. We use smaller units such as micro coulomb, etc.

## 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 F_{A} increases in magnitude while the force acting on B F_{B} 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 force F_{A} and F_{B} 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,

- It is used to calculate the distance and force between the two charges.
- It is used to arrange the charges in stable equilibrium.
- Columbus law is used to calculate electric field.

An electric field is given by,

E = F / Q_{T}(N/C)where,

is the Strength of the electric fieldEis the Electrostatic forceFQis the Test charge measured in coulombs_{T}

## Limitations of Coulomb’s Law

There are some limitations of Coulomb’s Law which are discussed below in the article,

- Coulomb’s Law is applicable for the point charges which are at rest.
- Coulomb’s Law is only applicable in situations where the inverse square law is followed.
- Coulomb’s Law is applicable only for the charges which are considered to be spherical. For charges with arbitrary shapes, Coulomb’s Law is not applicable because we cannot determine the distance between the charges.

**Also, Check**

## Solved Example on Coulomb’s Law

**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 charge at A and C are positive and the charge at B negative, what is the magnitude and direction of the total force on the charge at C?**

**Solution:**

Force F

_{CA}is applied toward AC and the expression for the F_{CA}is expressed as[Tex]F_{CA}=\dfrac{qq}{4\pi{\epsilon}_\circ} [/Tex]

Substitute the values in the above expression,

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

The Force F

_{CB}is applied toward CB and the expression for the F_{CB}is expressed as[Tex]F_{CB}=\dfrac{qq}{4\pi{\epsilon}_\circ} [/Tex]

Substitute the values in the above expression,

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

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,

[Tex]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} [/Tex]

**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 q

_{1}= 6×10^{-6}C.Second charge q

_{2}= 4×10^{-6}C.Distance between the charges r = 0.040 m

k = 9×10

^{9}We know that,

F = k q_{1}q_{2 }/ r^{2}Substitute the values in the above expression,

F = k q

_{1}q_{2 }/ r^{2}F = 9×10

^{9}×[(6×10^{-6})× (4×10^{-6})] / (0.04)^{2}F= 134.85 N

**Example 3:****Two-point charges, q**_{1}** = +9 μC and q**_{2}** = 4 μC are separated by a distance r = 12 cm. What is the magnitude of the electric force?**

**Solution:**

Given,

- k = 8.988 x 10
^{9}Nm^{2}C^{−2}- q
_{1}= +9μC = 9 × 10^{-6}C- q
_{2}= +4μC = 4 × 10^{-6}C- r = 12cm = 0.12m

F = k (q_{1}q_{2 }∕ r^{2})F = (8.9875 × 10

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

^{9 }) [36 × 10^{-12}/0.0144]

F = 22470 NElectric force between the charges is approximately

22.47 N

**For more ****Problems on Coulomb’s Law**

## FAQs on Coulomb’s Law

### Define Coulomb’s Law.

The electrostatic force of attraction or repulsion between two stationary point charges is directly proportional to the product of the magnitudes of the charges and inversely proportional to the square of the distance between them i.e.

F ∝ q_{1}q_{2}/ r^{2}

### State Coulomb’s Inverse-square Law in Electrostatics.

Coulomb’s Inverse-square law in electrostatics states that the force applied between two electrically charged particles is inversely proportional to the square of the distance between two particles.

### What is One Coulomb of Charge?

If one ampere of current passes through a conductor in one second then the charge transfer is one coulomb. One coulomb charge is also defined as the charge carried by 6×10

^{18}electrons.

### Is Electrostatic Force between any two point charges a central force?

Yes. The electrostatic force between two point charges always acts along the line joining the two charges. Hence it is a central force.

### What are Conditions for Stability of Coulomb’s Law?

Condition for stability of Coulomb’s Law is if a charge is introduced in a stable system of n charges then it is placed in such a position that the force on the test charge through all the other charges must cancel out each other.

### List some of the Application Of Coulomb’s Law.

Coulomb’s law is one of the fundamental laws of electrostatics and some applications of Coulomb’s Law are,

- It provides the force between electric charges.
- It is used to arrange electric charges in stable conditions.
- It is used to calculate the distance between two charges.
- It is used to tell the force acting at a point due to various charges.