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Why doesn't Gaus's law count for charges outside the area?

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  1. Feb 21, 2015 #1

    Jae

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    In Gaus's law when the integral is set up, we don't account for the charge outside the closed area. Why is this? How does this law work when the charges outside are not accounted for and only the charges enclosed is in the equation? I need an explanation why Gaus's law still works for calculating electric field when there are charges outside. This has been confusing me.

    Thank you for your time.
     
  2. jcsd
  3. Feb 21, 2015 #2
    The flux going into the surface from any outside charge necessarily exits the surface. Since Gauss's law takes into account the net flux through the surface, we can add up the individual contributions, and the net contribution from the outside charges will be 0.
     
  4. Feb 21, 2015 #3
    It may help to think about this in terms of flux. Imagine you have a single positive charge inside a closed surface. Its field lines extend radially and pass through the surface, so the flux is related to the charge inside. Field lines of a charge outside the surface do not contribute to the flux because its field lines go through one side and out the other, the surface isn't a closed surface for it. The left side of Gauss's law IS the equation for flux and therefore only pertains to charges within the closed surface, which is why it requires a surface integral.
     
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