.png){kind=link}
 - Vertical.png){kind=link}
.png){kind=link}
 - Vertical.png){kind=link}
.svg){kind=link}
.png){kind=link}
Coulomb's Law from Gauss's Law
\(\large ∯_A E \cdot dA = {\dfrac {q}{\varepsilon_{0}}}\) (Gauss's Law)
A uniform enclosure for a region would be a sphere.
For a sphere, surface area = \(\large 4 \pi r^2\)
Assuming the surface enclosing the charge is a sphere,
\(\large ∯_A E \cdot dA = E \cdot 4 \pi r^2\)
So,
\(\large E \cdot 4 \pi r^2 = {\dfrac {q}{\varepsilon_{0}}}\)
\(\implies \large E = {\dfrac{1}{4 \pi r^2}\dfrac {q}{\varepsilon_{0}}}\)
\(\implies \large E = {\dfrac{1}{4 \pi \varepsilon_{0}}\dfrac {q}{r^2}}\)