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I learned that the vector flux through any closed surface is always 0. So if you have a sphere(closed surface) and you put it in an uniform electric field, then the total flux is 0.

However, the Gauss's law states that the electric field flux through a closed surface equals the enclosed charge divided by the permitivity of free space. Shouldn't it be 0? What am I misunderstanding here?

Any explanation will be appreciated.

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    $\begingroup$ You learned wrong. The flux through an arbitrary closed surface is not zero. Gauss's law is correct. $\endgroup$ – garyp Apr 6 '16 at 13:53
  • $\begingroup$ @garyp I'm not saying it's wrong.. Then can you give me an example of a closed surface in an uniform electric field whose flux is not zero? $\endgroup$ – Hello Apr 6 '16 at 13:57
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    $\begingroup$ You have to think about what creates a "uniform electric field". If there was a charge anywhere in the uniform region, it would not be uniform anymore. $\endgroup$ – ACuriousMind Apr 6 '16 at 14:07
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    $\begingroup$ Example: A (gaussian) sphere between the plates of a large capacitor. The capacitor creates the uniform field. Total flux: zero. Now put a proton inside the sphere. Total flux: not zero. I'm saying that what you learned is wrong. The flux through any closed surface is not zero. $\endgroup$ – garyp Apr 6 '16 at 15:30
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A closed surface like a sphere encloses some volume. Anything coming out through the surface (the net outward flow which we call the flux) will be in the expense of what remains inside. If the sphere encloses some charge, then electric field diverging out from the volume containing the charge will be equal to the normal component of the electric field lines through the surface, which we call the electric flux.

The vector flux will be zero if the boundary and the surface are parallel. The electric filed is a special type of a vector which has a non-zero divergence if there is some non-zero charge. The electric flux will be zero only if there is no charge enclosing that surface.

However if you place an uncharged sphere in a uniform electric filed, the sphere develops induced charges. But there the charge is not residing inside the sphere but on the sphere. i.e, the charge induced is not enclosed by the sphere. So in that case the charge inside the sphere remains zero and you will get zero divergence and zero flux.

The Gauss's law states that the total electric flux coming out from a volume of charges is equal to the charge enclosed divided by the permittivity of the medium inside the surface.

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Not all vector fields have zero flux over any closed surface.

Those having zero flux everywhere are a special type called the 'solenoidal'. The electrostatic field doesn't belong to this group.

I think you need to study some vector analysis to capture the concepts correctly.

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  • $\begingroup$ what if you put it in an uniform electric field, will it still not have zero flux? $\endgroup$ – Hello Apr 6 '16 at 14:05
  • $\begingroup$ @LookAtTheBigPicture If an electric field was uniform over some region of space, then it has a zero flux over this region , but you can't get a uniform E-field in a region containing a net charge (you are not violating Gauss's law). $\endgroup$ – Tofi Apr 6 '16 at 14:25
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the electric flux linked wiith any closed surface of 0 charge is zero while the electri flux of a closed surface having some net charge is not zero say for example 2 hollow spheres one having a net charg n other having no charge ,the first will follow the gauss law while the other will follow the zero rule when the hollow sphere is kept in uniform electric field the field lines entering the hollow shere is equal to the field lines emerging out ,so it has a 0 net electric flux whereas a hollow sphere having a +ve charge will emit field lines,thus the elec5ric flux is not 0

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