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Confusion with Gauss's law .

  1. Jun 25, 2009 #1
    My physics textbook emphasizes that the electric field appearing in Gauss's law is the resultant electric field due to charges present both inside and outside the chosen closed surface , while the 'q' appearing in the law is only the charge contained within the surface. .This appears to follow from the mathematical statement of the law as the flux due to externally present charge is ,naturally , zero. But this also seems to suggest that the electric field on the Gaussian surface ( say a sphere ) would be the same whether there is solely one point charge ( which the sphere encloses say ) , or whether there are in addition a collection of point charges present outside the surface. But this cannot be true , can it ? Is there something I missed?
     
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  3. Jun 25, 2009 #2

    Doc Al

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    Even though an external charge doesn't change the net flux through the Gaussian surface, it still affects the field at a point on that surface.
     
  4. Jun 25, 2009 #3
    So , the electric field that is obtained from Gauss law is not the resultant electric field(due to enclosed charge + external charge) , but is the field due to the enclosed charge only. Am i right ?
     
  5. Jun 25, 2009 #4

    Born2bwire

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    You do not obtain the electric field from Gauss' Law, you obtain the net electric flux of a closed surface, which, as Doc Al notes, is unaffected by any charges that are not contained in the interior of the closed surface.
     
  6. Jun 25, 2009 #5

    Doc Al

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    Gauss's law tells you the net flux through the surface. For certain highly symmetric charge distributions, you can use the net flux to figure out the electric field. In those special situations the electric field you find is the total field due to all charges, but the contribution to the field from external charges will be zero.
     
  7. Jun 25, 2009 #6
    Understood. But what if the external charges modify the field at the Gaussian surface. Doesn't Gauss's law fail to take them into account , simply because they do not contribute to the net flux through the closed surface.
     
  8. Jun 25, 2009 #7

    Doc Al

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    If the external charges modify the field at the Gaussian surface, then the situation lacks sufficient symmetry to use the net flux to calculate the field.

    Gauss's law always gives you the correct net flux, but only in special cases can it be used to find the field.
     
  9. Jun 25, 2009 #8
    Okay ......I get it now. Thanks
     
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