Question

WHY DISTANCE IS CONSIDERED SQUARE IN COULOMB’S LAW

Answer 1

Answering your question with analogy to Newton's Law of Gravitation,

Consider a charged object moving in a circle with constant speed,

$F=centripetal\space force$

$F=\dfrac{mv^2}{r}$

For an object moving in circle,

$v=\dfrac{2\pi r}{T}$

Substituting the value of $v$ in above equation

$F=\dfrac{m(\dfrac{2\pi r}{T})^2}{r}$

$F=\dfrac{4m\pi^2r}{T^2}$

$F=4m\pi^2\dfrac{r}{T^2}$

From Kepler's third law,

$T^2\propto r^3$

$T^2=kr^3$

Substituting the value of $T^2$ , in above equation,

$F=4m\pi^2\dfrac{r}{kr^3}$

$F=\dfrac{4m\pi^2}{k}\times\dfrac{1}{r^2}$

$F\propto\dfrac{1}{r^2}$