To derive Coulomb’s law from Gauss’s law, we first need to understand Coulomb’s law, Gauss’s law, and Electric flux.

**Coulomb’s law:** According to this law, the magnitude of the force between two point charges separated by distance r in air or vacuum is directly proportional to the product of magnitudes of these charges and inversely proportional to the square of the distance between them.

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

**Electric flux:** The electric flux (Φ_{E}) through any surface is defined as the total number of electric field lines passing through that surface.

$$ Electric\: flux, \:\phi_E = \oint \vec E . \vec {ds}$$

**Gauss’s law:** According to Gauss’s law, the total electric flux (Φ_{E}) passing normally through a surface in free space or vacuum is equal to 1/ε_{0} times the total electric charge (q) enclosed by the surface.

$$ \phi_E = \frac {q}{\epsilon_0} \Rightarrow \oint \vec E . \vec {dS} = \frac {q}{\epsilon_0} $$

Where ε_{0 }is the permittivity of free space or vacuum.

## Derivation of Coulomb’s law from Gauss’s law

**To Prove:** Coulomb’s law i.e. the electric force between two charges q_1 and q_2, using Gauss’s law.

**Proof: **Consider two point charges q_{1} and q_{2} separated by distance r, are placed in free space. Let E be the intensity of the electric field at distance r due to charge q_{1}. Then the force experienced by charge q_{2} due to electric field E is given by

\( F = q_2E\) …(1)

Now, we will draw a Gaussian sphere of radius r taking the position of charge q_{1} as the center of the sphere.

According to Gauss’s law, total electric flux through this closed Gaussian sphere is given by

\(\oint \vec E . \vec {dS} = \frac {q_1}{\epsilon_0}\)

\(\Rightarrow \oint E \:dS \: cos \theta = \frac {q_1}{\epsilon_0}\)

Here dS is the small area element of the Gaussian sphere and θ is the angle between the direction of the electric field and the area vector i.e. θ = 0^{°}.

\(\Rightarrow \oint E \:dS \: cos \:0 = \frac {q_1}{\epsilon_0}\)

\(\Rightarrow \oint E \:dS = \frac {q_1}{\epsilon_0}\)

Since the magnitude of electric field intensity is constant at every point of the Gaussian sphere, therefore we can write

\( E \oint dS = \frac {q_1}{\epsilon_0}\) …(2)

But, \( \oint dS \) is the surface area of the Gaussian sphere, which is given by

\( \oint dS = 4 \pi r^2 \)

Hence, equation (2) becomes

\( E \times 4 \pi r^2 = \frac {q_1}{\epsilon_0}\)

\( \Rightarrow E = \frac {1}{4 \pi \epsilon_0} \frac {q_1}{r^2}\)

Substituting the above expression in equation (1), we get

\( F = q_2 \times \frac {1}{4 \pi \epsilon_0} \frac {q_1}{r^2} \)

**\( \Rightarrow F = \frac {1}{4 \pi \epsilon_0} \frac {q_1 q_2}{r^2}\)**

The above expression is known as Coulomb’s law.