Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

In electrostatics total flux linked from the closed surface enclosing the charge is equal to $Q/\varepsilon_0$. This is according to Gauss Law. Is this the experimental value or defined value. If experimental what type of experiment is done by Gauss? Why this value is chosen?

share|improve this question

2 Answers 2

up vote 1 down vote accepted

Let's derive Gauss' Law

enter image description here

Let's take a small area surface on the gaussian surface $\Delta S$ and let's calculate flux through it.And then integrate over complete surface.

$$\phi = \int_s d\phi =\int_s E \Delta S$$

$$\phi=\int_s \dfrac q{4\pi\epsilon_0 r^2} \Delta S$$

and $\dfrac{\Delta S}{r^2} =\Delta \Omega $ (solid angle).

$$\phi = \dfrac{q}{4\pi\epsilon_0} \int_s\Delta \Omega$$

$$\phi = \dfrac{q}{4\pi\epsilon_0} 4\pi=\dfrac q{\epsilon_0}$$

share|improve this answer

This can be verified theoretically. As the electrostatic force (Coulomb force) is a central force ,i.e. there is spherical symmetry we can use Gauss's divergence law easily. At this the constant $\epsilon_0$ emerges automatically. The value of this constant depends upon the medium. For different media it can be experimentally measured. As example using two concentric hollow cylinders one can measure $\epsilon$ for any liquid. The Coulomb law is an empirical law.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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