Derive Coulomb’s law from Gauss’s law

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 (ΦE) 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 (ΦE) passing normally through a surface in free space or vacuum is equal to 1/ε0 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 εis 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 q1 and q2 separated by distance r, are placed in free space. Let E be the intensity of the electric field at distance r due to charge q1. Then the force experienced by charge q2 due to electric field E is given by

\( F = q_2E\)    …(1)

Force experienced by a charge in electric field
Force experienced by a charge in an electric field

Now, we will draw a Gaussian sphere of radius r taking the position of charge q1 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°.

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

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

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

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.

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