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Why is it that objects kept in uniform electric field have net flux through them when gauss law says that net flux through a surface is q/epsilon.I mean the net charge will be zero if a gausian surface is drawn so the net flux must also be the same

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Gauss's law says that the net flux through a closed surface is equal to $q/\epsilon_0$ where $q$ is the charge enclosed in the surface. Gauss's law would not make sense for open surfaces that don't enclose a volume. For a uniform electric field, any closed surface will have equal number of field lines entering the surface as exiting the surface and so the total flux across that surface will be 0 as required.

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  • $\begingroup$ then why is it that there is a net flux through a disc kept in uniform electric field? $\endgroup$ – chemophilic Sep 27 '18 at 20:15
  • $\begingroup$ @MohdKhan Why wouldn't there be? There is a net flux through the disc (assuming it's not being aligned parallel to the field lines), but the disc is not a closed surface and so you can't apply Gauss's law to it. $\endgroup$ – enumaris Sep 27 '18 at 20:17
  • $\begingroup$ As i have been told gaussian surfaces are hypothetical and need not 'come with the body' and can be drawn as per one's convinience.So what if i draw a hypothetical gausian surface enclosing the disc $\endgroup$ – chemophilic Sep 27 '18 at 20:26
  • $\begingroup$ 1 more doubt..As per what you have said if the disc has to be a closed surface it becomes a volume..But gauus's law talks about gaussian surfaces $\endgroup$ – chemophilic Sep 27 '18 at 20:34
  • $\begingroup$ Gaussian surfaces need to be closed, a disc is not closed. If you drew a closed Gaussian surface anywhere (whether it encloses a hypothetical disc or not) you will get a net flux of 0 across the closed surface. Perhaps take a look here: en.wikipedia.org/wiki/Gaussian_surface to see an explanation of what a Gaussian surface is. $\endgroup$ – enumaris Sep 27 '18 at 20:35

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