Coulomb’s Law – Intuitive, Mathematical & Exam-Oriented Explanation

1. Introduction: Why Coulomb’s Law Matters So Much

Coulomb’s Law is the first fundamental law of Electrostatics.
Every concept that follows—electric field, potential, Gauss’s law, capacitors—
is built on this single idea.

2. What Is Coulomb’s Law? (Conceptual Meaning)

Coulomb’s Law describes the electrostatic force between two point charges at rest.

  • Like charges → Repulsion
  • Unlike charges → Attraction

3. Historical Background

Charles-Augustin de Coulomb established this law using a torsion balance,
showing force depends on charge and distance.

4. Coulomb’s Law – Mathematical Form

Scalar Form

F = k × ( q₁ q₂ ) / r²
  • F = electrostatic force
  • q₁, q₂ = charges
  • r = distance
  • k = 1 / (4πϵ₀)
k = 9 × 10⁹ N m² C⁻²

Vector Form (IIT-JEE & IB)

F⃗₁₂ = (1 / 4πϵ₀) × ( q₁ q₂ / r² ) r̂₁₂

5. Diagram Explanation

For like charges, force arrows point away.
For unlike charges, arrows point toward each other.

6. Factors Affecting Electrostatic Force

(a) Magnitude of Charges

Force ∝ q₁q₂

(b) Distance

Force ∝ 1 / r²

(c) Medium

F = (1 / 4πϵ₀ϵᵣ) × ( q₁ q₂ / r² )
Medium Relative Permittivity
Vacuum 1
Air ~1
Water ~80

7. Physical Interpretation

  • Long-range force
  • Central force
  • Pairwise interaction

8. Limitations of Coulomb’s Law

  • Valid for point charges
  • Charges must be at rest
  • Fails at relativistic speeds

9. Coulomb’s Law vs Newton’s Law

Feature Gravitational Electrostatic
Nature Attractive Attractive / Repulsive
Strength Weak Very Strong

Electrostatic force is 10³⁶ times stronger.

10. Superposition Principle

Net force equals vector sum of individual forces.

11. Common Student Mistakes

  • Wrong force direction
  • Ignoring medium
  • Vector mistakes

12–14. Exam-Oriented Questions

CBSE, NEET, IIT-JEE & IB questions test concepts + vectors.

15. Why Students Struggle

Formula memorisation without understanding causes fear.

16. Right Way to Master Coulomb’s Law

  • Understand interaction
  • Draw diagrams
  • Use vectors
  • Apply superposition

17. Connection to Next Topics

  • Electric Field
  • Potential
  • Gauss’s Law
  • Capacitors

18. Personal Guidance with Kumar Sir

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19. Final Takeaway

Coulomb’s Law becomes easy when learned conceptually.
Once physics is clear, mathematics follows naturally.

Electric Dipole and Dipole Moment

Definition: An electric dipole is a system consisting of two equal and opposite point charges ($+q$ and $-q$) separated by a very small distance ($2a$).

Standard Dipole Diagram

O a a -q +q Dipole Moment (p)

Fig: Geometry of an electric dipole with center O and separation 2a.

[Image of electric dipole and dipole moment]

1. Electric Dipole Moment ($\vec{p}$)

It is a vector quantity that measures the strength of the dipole. It is defined as the product of the magnitude of either charge and the distance between them.

$$ \vec{p} = q \times (2\vec{a}) $$

Key Features for JEE/NEET:

  • Direction: By convention, it is always from Negative charge ($-q$) to Positive charge ($+q$).
  • SI Unit: Coulomb-meter ($C \cdot m$).
  • Dimensions: $[L T A]$.
  • Ideal Dipole: When charge $q \to \infty$ and separation $2a \to 0$ such that the product $p = q(2a)$ remains finite.

2. Electric Field due to Dipole

Position Formula (General) Short Dipole ($r \gg a$)
Axial Point $$ \frac{1}{4\pi\epsilon_0} \frac{2pr}{(r^2-a^2)^2} $$ $$ \frac{1}{4\pi\epsilon_0} \frac{2p}{r^3} $$
Equatorial Point $$ \frac{1}{4\pi\epsilon_0} \frac{p}{(r^2+a^2)^{3/2}} $$ $$ \frac{1}{4\pi\epsilon_0} \frac{p}{r^3} $$

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Coulomb’s Law in Vector Form

F₁₂ F₂₁ q₁ q₂ r

Fig: Repulsive forces between two like charges $q_1$ and $q_2$

1. Mathematical Formula

$$\vec{F}_{21} = \frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{r^2} \hat{r}_{21}$$

2. Important Highlights

  • Newton’s Third Law: The force exerted by $q_1$ on $q_2$ is equal and opposite to force by $q_2$ on $q_1$: $\vec{F}_{12} = -\vec{F}_{21}$.
  • Central Force: Forces act along the line joining the centers of the two charges.

Prepared for physicsclass.com

Electric Field at an Axial Point of a Dipole

Definition: An axial point is a point lying on the line joining the two charges of the electric dipole. We aim to find the net electric field at point P located at distance r from the center of the dipole.

Vector Diagram: Axial Line

O -q A +q B P E₊ E₋

Fig: Electric field vectors $\vec{E}_+$ and $\vec{E}_-$ at observation point P.

Step-by-Step Derivation

1. Field due to $+q$ at P:
Distance $BP = (r – a)$. Therefore, $E_+ = \frac{1}{4\pi\epsilon_0} \frac{q}{(r-a)^2}$ (Away from dipole)

2. Field due to $-q$ at P:
Distance $AP = (r + a)$. Therefore, $E_- = \frac{1}{4\pi\epsilon_0} \frac{q}{(r+a)^2}$ (Towards dipole)

3. Net Electric Field ($E_{net}$):
Since $E_+ > E_-$, the net field is $E_{axial} = E_+ – E_-$.

$$ E_{axial} = \frac{q}{4\pi\epsilon_0} \left[ \frac{1}{(r-a)^2} – \frac{1}{(r+a)^2} \right] $$

Solving the brackets…

$$ E_{axial} = \frac{1}{4\pi\epsilon_0} \frac{q \cdot 4ra}{(r^2-a^2)^2} $$

Final Result

Since Dipole Moment $p = q \times 2a$, we get:

$$ \vec{E}_{axial} = \frac{1}{4\pi\epsilon_0} \frac{2\vec{p}r}{(r^2 – a^2)^2} $$

Special Case: For a short dipole ($r >> a$), the formula simplifies to:

$$ E_{axial} \approx \frac{1}{4\pi\epsilon_0} \frac{2p}{r^3} $$

Key Takeaway: The direction of the electric field at any axial point is always same as the direction of the dipole moment ($\vec{p}$).

Electric Field at an Equatorial Point

Definition: An equatorial point is a point lying on the perpendicular bisector of the dipole. We aim to find the field at point P at a distance r from the center O.

Vector Resolution Diagram

r -q +q O a a P E₊ E₋ E_net

Fig: Resolving $E_+$ and $E_-$ into components. The sine components cancel out.

Step-by-Step Derivation

1. Individual Magnitudes:
The distance of point P from either charge is $x = \sqrt{r^2 + a^2}$.
$|E_+| = |E_-| = \frac{1}{4\pi\epsilon_0} \frac{q}{(r^2 + a^2)}$

2. Resolving Components:
The vertical components ($E_+ \sin\theta$ and $E_- \sin\theta$) are equal and opposite, so they cancel out. The horizontal components ($E_+ \cos\theta$ and $E_- \cos\theta$) act in the same direction and add up.

$$ E_{net} = E_+ \cos\theta + E_- \cos\theta = 2 E_+ \cos\theta $$

From the geometry, $\cos\theta = \frac{a}{\sqrt{r^2+a^2}}$.

$$ E_{equi} = 2 \left( \frac{1}{4\pi\epsilon_0} \frac{q}{(r^2 + a^2)} \right) \frac{a}{(r^2 + a^2)^{1/2}} $$

Final Vector Expression

Using $p = q \times 2a$ and noting that the field is opposite to the direction of $\vec{p}$:

$$ \vec{E}_{equi} = – \frac{1}{4\pi\epsilon_0} \frac{\vec{p}}{(r^2 + a^2)^{3/2}} $$

Short Dipole Case ($r \gg a$):

$$ E_{equi} \approx \frac{1}{4\pi\epsilon_0} \frac{p}{r^3} $$
Important Comparison: For the same distance $r$, the axial field is twice the equatorial field:
$E_{axial} = 2 \times E_{equi}$

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