Derive coulomb's law from gauss's law

According to gauss's law, the total electric flux (ϕ) through any close surface in free space is equal to 1/∈₀ times the total electric charge (q) enclosed by the surface.

Mathematically,

\(\phi = \oint _S \overrightarrow{E} . \overrightarrow{ds} = \frac{q}{ \in _0}\)

According to coulomb’s law, the magnitude of the force of attraction and repulsion between any two point charges at rest is directly proportional to the product of the magnitude of charges and inversely proportional to the square of the distance between them.

Mathematically, it is given by:

\(F=k \frac{q_1q_2}{r_2}\)

Derivation of coulomb’s law from gauss’s law

Consider two point charges q₁ and q₂ are separated by distance r in a vacuum.

Let E be the magnitude of the electric field at distance r from charge q₁. Therefore, the force(F) experienced by charge q₂ due to electric field(E) is given by:

F = q₂E …(i)

Draw a gaussian surface of radius r taking the center as the location of charge q₁.

The intensity of the electric field is constant at every point of this surface.

According to gauss’s law, the net electric flux through this surface is given by

\(\oint _S \overrightarrow{E} . \overrightarrow{ds} = \frac{q_1}{ \in _0}\)

⇒ \(\oint _S\) Eds cosθ = \(\frac{q_1}{ \in _0}\)

Direction of the electric field is always perpendicular to the surface. So, the angle between E and ds is 0.

⇒ \(\oint _S\)Eds cos0 = \(\frac{q_1}{ \in _0}\)

⇒ \(\oint _S\) Eds = \(\frac{q_1}{ \in _0}\)

Since the electric field is constant at every point of the gaussian surface, therefore we can write

E \(\oint _S\) ds = \(\frac{q_1}{ \in _0}\)

But, \(\oint _S\) ds = 4πr² is the surface area of sphere

⇒ E x 4πr² = \(\frac{q_1}{ \in _0}\)

Using equation (i), we get

\(\frac{F}{q_2}\) x 4πr²  = \(\frac{q_1}{ \in _0}\)

⇒ F = \(\frac{1}{4 \pi \in _o} \frac{q_1q_2}{r_2}\)

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