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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?

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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).

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

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

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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.

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