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We define flux of an electric field $\vec{E}$ through a surface $S$ as

$$\Phi =\int_S\vec{E}.d\vec{s}$$

Now we define flux through a surface as stated above

What does it mean when we say flux coming out of a $q$ charge is $\displaystyle\frac{q}{\epsilon_0}$ ?

I think this statement is not correct because flux is not something which comes out of a charge and even if it would have been like that why would it will always be $\displaystyle\frac{q}{\epsilon_0}$ until we have a closed surface around it rather we have electric lines of forces which can originate from a charge.

The doubt arise when my teacher said that flux of arbitrary closed surface having $q$ charge inside it is $\displaystyle\frac{q}{\epsilon_0}$ because flux originating/emitting from the $q$ charge is $\displaystyle\frac{q}{\epsilon_0}$.

The correct statement should have been flux of a spherical closed surface having $q$ charge inside it is $\displaystyle\frac{q}{\epsilon_0}$ and since relative number of field lines originating from $q$ charge is same for both spherical surface and arbitrary surface , therefore flux through both is same

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The correct statement would be "The flux through the closed surface created(due to) by a charge $q$ inside is $\frac{q}{\epsilon _0}$". The word used should be through because flux is directly proportional to the number of electric field lines passing through it.

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  • $\begingroup$ Yes I also think the same. Thankyou for your answer $\endgroup$
    – Lalit Tolani
    Sep 18, 2021 at 15:38
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It is intuitively useful to think of flux being proportional to the number of field lines passing through a surface, but since field lines don't actually exist, a more rigorous prof that the total flux of field coming from a point charge is independent of shape or size of the surrounding surface, generally involves applying the dot product in the definition of flux to the outer surface of a very small solid angle. (The field decreases with the square of the distance, and the radial component of the area vector increases with the square of the distance.)

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  • $\begingroup$ Yes I know this but what does it mean when we say the flux emitting from a charge $\displaystyle\frac{q}{\epsilon_0}$ $\endgroup$
    – Lalit Tolani
    Sep 18, 2021 at 15:37
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What does it mean when we say flux coming out of a $q$ charge is $\displaystyle\frac{q}{\epsilon_0}$ ?

It is not flux coming out of a charge. Flux is due to the electric field produced by charges. Moreover, Gauss' law says that the net flux (not just the flux) over a closed surface equals the charge enclosed divided by the permittivity of the space.

why would it will always be $\displaystyle\frac{q}{\epsilon_0}$ until we have a closed surface around it rather we have electric lines of forces which can originate from a charge.

It has to be a closed surface in order to apply Gauss' law. In terms of field lines, every field line originating from a charge outside the enclosed surface that enters the enclosed volume across some part of the surface, has to exit the enclosed volume at some other part of the surface, for a net flux of zero. If the surface were "open" in some area, the next flux due to the field of external charges may not be zero.

The correct statement should have been flux of a spherical closed surface having $q$ charge inside it is $\displaystyle\frac{q}{\epsilon_0}$ and since relative number of field lines originating from $q$ charge is same for both spherical surface and arbitrary surface , therefore flux through both is same

Not sure exactly all your teacher told you, but yes Gauss's law applies to any any arbitrary closed surface. Again, it is important that you use the term net flux, because the law only addresses the net flux (field lines) across the surface, not the distribution of the flux (field lines) across the surface.

Hope this helps.

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