2
$\begingroup$

I read about Coulomb's law it says the force of attraction or repulsion between two charges is inversely proportional to the square of the distance between them.

So generally if I say then Coulomb's law is inverse square law. But I really want to know why this electrostatic force depends on square of the distance, and not the cube nor the fourth power. Why only the square of the distance?

Is there any connection with dimensions? Or does it depend on the reference frame? Why does it depend on inverse of the square of the distance?

$\endgroup$

marked as duplicate by Qmechanic Nov 2 '16 at 11:49

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ Heard of Gauss' Law? $\endgroup$ – user36790 Nov 1 '16 at 19:15
  • $\begingroup$ While there are some deeper backdrops for this one, at some point the answer to every "why?" question in physics is simply "Because.". We describe how the universe works, and anything else is a bonus extra. $\endgroup$ – Emilio Pisanty Nov 1 '16 at 19:28
  • 1
    $\begingroup$ Possible duplicates: physics.stackexchange.com/q/93/50583, physics.stackexchange.com/q/47084/50583 $\endgroup$ – ACuriousMind Nov 1 '16 at 19:28
  • 1
    $\begingroup$ "Why does it depend on inverse of the square of the distance?" - in the context of quantum field theory, the answer might be that the photon is massless. See, for example, Yukawa potential $\endgroup$ – Alfred Centauri Nov 1 '16 at 19:59
4
$\begingroup$

It's just because of "Gauss's law for Electric fields(Maxwell's first equation)"

Taking S in the integral form of Gauss's law to be a spherical surface of radius r, centered at the point charge Q, we have $${\displaystyle \oint _{S}\mathbf {E} \cdot d\mathbf {A} ={\frac {Q}{\varepsilon _{0}}}}$$ By the assumption of spherical symmetry, the integrand is a constant which can be taken out of the integral. The result is $${\displaystyle 4\pi r^{2}{\hat {\mathbf {r} }}\cdot \mathbf {E} (\mathbf {r} )={\frac {Q}{\varepsilon _{0}}}}$$ where ${\displaystyle {\hat {\mathbf {r} }}}$ is a unit vector pointing radially away from the charge. Again by spherical symmetry, E points in the radial direction, and so we get $${\displaystyle \mathbf {E} (\mathbf {r} )={\frac {Q}{4\pi \varepsilon _{0}}}{\frac {\hat {\mathbf {r} }}{r^{2}}}}$$ which is essentially equivalent to Coulomb's law. Thus the inverse-square law dependence of the electric field in Coulomb's law follows from Gauss's law.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.