Coulomb's Law Class 12: Detailed for CBSE Exams and Beyond

Coulomb's Law explains the formula for electrostatic force between two charges. These are two point charges q_1 and q_2, which are separated by a distance r in a vacuum.

“Coulomb’s law is a quantitative statement about the force between two point charges.”
You learn this statement first in Class 12 Physics NCERT textbook in section 1.5. 

NCERT further explains that
“Coulomb measured the force between two point charges and found that it varied inversely as the square of the distance between the charges and was directly proportional to the product of the magnitude of the two charges and acted along the line joining the two charges. Thus, if two point charges q1, q2 are separated by a distance r in vacuum,the magnitude of the force (F) between them is given by

$F=k\frac{\left|q_1{q}_2\right|}{r^2}$ "

Let's see more of how the maths or what Coulomb's Law really is  

 F ∝ q_1.q₂ $$F = k \frac{|q_1 q_2|}{r^2}$$

Combine the two and you get the equation below. 

$F=k\frac{\left|q_1{q}_2\right|}{r^2}$

 

This is the Coulomb's Law formula. Here, 

• $F$is the magnitude of electrostatic force, which is N

• $q_1, q_2$are the charges in coulombs that we represent as C

• $r$refers to the distance between charges in metres (m)

• $k$is the Coulomb constant that has the equation like
  $\frac{1}{4\pi \varepsilon_0}$

• $\varepsilon_0$ is the permittivity of free space and is $8.854 \times 10^{-12} \, \text{C}^2/\text{N·m}^2$
  

• $k$in vacuum is $9 \times 10^9 \, \text{N·m}^2/\text{C}^2$

The actual formula of Coulomb's law, Electrostatic Force in a medium with relative permittivity $\varepsilon_r$:

$$F=\frac{1}{4\pi {\epsilon }_0{\epsilon }_r}\frac{\mathrm{\mid }{q}_1{q}_2\mathrm{\mid }}{r^{\mathrm{2 }}}$$

Related Topics: Class 12 Physics Chapter 1 NCERT Solutions

The force between the charges is directly proportional to the product of the charges. $F\propto \; <!--\; \propto \; -->q_1{q}_2$

The electrostatic force, as Coulomb's Law mentions that it's inversely proportional to the square of the distance between the charges. $F\propto \; <!--\; \propto \; -->\frac{1}{r^2}$

 

The equation we get is $F\propto \; <!--\; \propto \; -->(q_1{q}_2)/r^2$

After we remove the proportional sign, a constant is introduced, which is Coulomb's Constant.

This constant value is based on the medium's permittivity.

Coulomb's Constant (k) is expressed as  $k=\frac{1}{4\mathrm{\pi \; <!--\; \pi \; -->}{\mathrm{ϵ\; <!--\; ϵ\; -->}}_o}$

Where, 

${\mathrm{ϵ\; <!--\; ϵ\; -->}}_o$ is the permittivity of vacuum. 

In any medium, the Coulomb's constant will be 

$k=\frac{1}{4\mathrm{\pi \; <!--\; \pi \; -->}{\mathrm{ϵ\; <!--\; ϵ\; -->}}_r{\mathrm{ϵ\; <!--\; ϵ\; -->}}_o}=k=\frac{1}{4\mathrm{\pi \; <!--\; \pi \; -->}\mathrm{ϵ\; <!--\; ϵ\; -->}}$

Where, 

${\mathrm{ϵ\; <!--\; ϵ\; -->}}_r$ is the medium's permittivity 

$\mathrm{ϵ\; <!--\; ϵ\; -->}$ is the permittivity of the medium with respect to vacuum 

Coulomb's Law provides the formula used to determine the electrostatic force. This law applies specifically to two point charges, q₁ and q₂, that are separated by a distance, r, within a vacuum.

$$F = k\,\frac{|q_1 q_2|}{r^2},\quad k=\frac{1}{4\pi\varepsilon_0}.$$

This formula for Coulomb's Law Class 12 tells us only the magnitude of the force. 

The vector form also needs to show us where the force is exerted, ie., the direction.

Now, let's assume the displacement vectors:

$${\mathbf{r}}_{12}={\mathbf{r}}_1-{\mathbf{r}}_2(\text{from point 2 to 1)}{\mathrm{<br>}}_{}<br>$$

$${\mathbf{r}}_{21}={\mathbf{r}}_2-{\mathbf{r}}_1=-{\mathbf{r}}_{12}\mathrm{ \; (from\; point\; 1\; to\; 2)}$$

Their magnitude will be $r = \lVert \mathbf{r}_{12} \rVert = \lVert \mathbf{r}_{21} \rVert$.

The unit vectors along the line that joins the charges

$$\hat{\mathbf{r}}_{12} = \frac{\mathbf{r}_{12}}{\lVert \mathbf{r}_{12} \rVert},\qquad \hat{\mathbf{r}}_{21} = \frac{\mathbf{r}_{21}}{\lVert \mathbf{r}_{21} \rVert}=-\hat{\mathbf{r}}_{12}.$$

So the Electrostatic Force on $q_1$q1​ due to $q_2$q2​ in vector form:

$$\boxed{\;\mathbf{F}_{12} \;=\; k\,\frac{q_1 q_2}{\lVert \mathbf{r}_{12} \rVert^{2}}\,\hat{\mathbf{r}}_{12}\;}$$

Actual Vector Form of Coulomb's Law (with no unit vector):

$$\boxed{\;\mathbf{F}_{12} \;=\; k\,\frac{q_1 q_2}{\lVert \mathbf{r}_{12} \rVert^{3}}\,\mathbf{r}_{12}\;}$$

Since $\hat{\mathbf{r}}_{12}=\mathbf{r}_{12}/\lVert\mathbf{r}_{12}\rVert$.

How Coulomb's Law and Newton's Third Law Connect

We will show how Coulomb's Law, with the unit vector and force vector components in the electrostatic force formula, has a direct link to Newton's Third Law of Motion. 

Unit vector in the Electrostatic Force Formula: The unit vector r̂₂₁ defines the direction from point charge 1 to 2. We have to use this vector as the arrow, which is pointing from the source of the charge (1) to the target source (2).

Force Vector in the Electrostatic Force Formula: F_12 is the symbol which shows the force from q_1 acting on q_2. This force has a strength, which is the other term for magnitude, and which is calculated using Coulomb's Law. The direction is determined by the unit vector.

Forces of Action-Reaction Pairs in Newton's Third Law Explains Coulomb's Law

We have previously learnt in Class 11 Physics about Newton's Third Law that for all actions, there will be reactions that are equal and opposite. In the Coulomb's Law context, when we apply this law, we can say that, when the charge, q_1 creates an action by exerting some force on q_2, we can deduce that q_2 will also be exerting an opposite force on q_1. 

So the force we can say to be 

F_12 = - F_21

Now that tells us the two forces have equal sizes, they are opposite in direction, and lie along the same line of charges that join them. 

Until now, we have been focused on the electrostatic force between two point charges to understand Coulomb’s Law. 

What would happen to it if there were multiple point charges?

You see, when there are electrostatic forces from various other stationary charges on one charge, the net force that acts on it is the vector sum of all the individual forces.

${\stackrel{\to }{F}}_{\mathrm{net}}=\underset{i=1}{\overset{n}{\sum }}{\stackrel{\to }{F}}_i$

One thing to learn about the superposition principle here is that we do not have to consider the force between any two charges. This one force is practically unaffected by other charges present in the system. To calculate the net force, we need to use vector addition rules like the parallelogram and triangle laws to get results on charge assemblies accurately. 

We cannot apply Coulomb’s Law for all charges and forces. It is applicable to point charges that remain stationary. Other than that, the charges have to be distributed in such a way that they are spherically symmetrical. One example would be metal spheres that are uniformly charged. Now, if it’s an irregular shape, the law won’t hold. 

The second major limitation of Coulomb’s Law is that it assumes charges do not move relative to one another. That electrostatic condition is not possible for charges that are in motion. The law does not account for magnetic effects and the finite speed of moving interactions. 

Coulomb’s Law relies on the inverse square relationship between force and distance. If there is any deviation from this relationship, the accuracy of the results becomes limited. 

A few things to remember when looking at the difference between Coulomb's force and gravitational force. 

Property	Electrostatic Force (F_e)	Gravitational Force (F_g)
Nature of Force	It can be attractive or repulsive	Always has to be attractive.
Constant	k = 8.99 × 10⁹ N·m²/C²	G = 6.67 × 10⁻¹¹ N·m²/kg²
Operating Charges	Electric charges (q)	Masses (m)
Direction	Along line joining charges and only depends on charge signs	Along line joining masses and always attractive
Relative Strength	F_e/F_g ≈ 2.27 × 10³⁹, which is the proton-electron pair	It’s much too weak to appear negligible at atomic scales
Large-scale Effect	This cancels out in electrically neutral bodies	It only dominates planetary and cosmic interactions

For gaining mastery over Coulomb’s Law for CBSE boards, have a look at these question types and approaches.

Type of Exam Question on Coulomb’s Law	What to Use	What Concept to Remember
For directly calculating	Simply use F = k|q₁q₂|/r²	Check for how unit conversions happen and where the formula applies
All inverse problems	Try to rearrange the Coulomb’s Law formula to find q or r	It’s a simple change in the algebra of Coulomb's law
Superposition principle with more than two charges	Look for the net force in multi-charge setups	Vector addition formulas apply here