Doubt in the derivation of Gauss's law in dielectrics

Question

so in the 2nd page,when the dielectric material is introduced the gauss's law becomes  $$oint _ { S } vec { E } cdot vec { d S } = frac { ( q - q _ { i } ) } { epsilon _ { 0 } }$$.But my question is why the ${ epsilon _ { 0 } }$is in the equation.Shouldn't it be  ${ epsilon }(varepsilon = k varepsilon _ { 0 })$ ?And the formula becomes

$$oint _ { S } vec { E } cdot vec { d S } = frac { ( q - q _ { i } ) } { k varepsilon _ { 0 } }$$

Because

${ epsilon _ { 0 } }$ is used when the medium is air or vacuum, but here the medium is dielectric near the gaussian surface, so ${ epsilon }$  should be used instead of ${ epsilon _ { 0 } }$ in the gauss's law here.

Puk · Accepted Answer

The book is correct, although this is not quite a general derivation since it derives Gauss' law for a specific structure, and assumes a linear and uniform dielectric.
Gauss' law in the form
$$ ointlimits_S{vec{E}·dvec{S}} = frac{q}{epsilon_0}. $$
is always valid, dielectric or not. $q$ here is the total charge enclosed by $S$, including charges induced in dielectrics. The alternative form of Gauss' law, sometimes called the macroscopic form, is
$$ ointlimits_S{vec{D}·dvec{S}} = q_f $$
where $vec{D}$ is the electric flux density equal to $epsilon vec{E}$ in linear media, and $q_f$ is the free charge enclosed by $S$, not including induced charges. In a uniform linear medium, this equation can also be written as
$$ ointlimits_S{vec{E}·dvec{S}} = frac{q_f}{epsilon} = frac{q_f}{kappaepsilon_0}. $$

Doubt in the derivation of Gauss's law in dielectrics

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