Will an infinite plane of charge flux generate flux through itself?

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SUMMARY

An infinite plane of charge does not generate electric flux through itself when a hole is cut into it. The electric field (E-field) within the hole remains zero due to symmetry, confirming that no flux enters or exits the hole. In scenarios where the infinite plate is part of a closed conducting surface, the E-field in the hole is E_0/2. This conclusion is derived from applying Gauss's law and symmetry arguments in electrostatics.

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  • Understanding of Gauss's Law in electrostatics
  • Familiarity with electric fields and flux concepts
  • Knowledge of symmetry arguments in physics
  • Basic principles of conducting surfaces in electrostatics
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ozone
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I just have a quick question, and I'm guessing the answer is no but I wanted to make sure that this was sensible. In general whenever we consider flux we think of some kind of closed surface or a scenario where charge closes back on itself.

If I were to cut a hole in an infinite plate of charge, would there be any flux through that hole from the plate of charge? My guess is that we can take the limiting case as a pillbox on the hole shrinks, and we will see that no flux is entering or exiting (I think the net E-field ought to be 0 in the hole). However it doesn't make sense to me why anyone would ask this question then.
 
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The electric field in the hole must be zero, yes. You can see that by symmetry arguments alone. Why would anybody ask that question? Because they want to know the answer, and so did you...
 
If the plate is part of a closed conducting surface, the E field in the hole will be E_0/2.
 

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