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Gauss Law of Cube in non-uniform linear Electric Field.

  1. Dec 9, 2014 #1
    Gauss's Law states that if a Gaussian Surface encloses a charge ##q_{enc}## then the electric flux through the Gaussian Surface is given by ##\phi=q_{enc}/\varepsilon_{o}## .

    It also states that any external field does not contribute to the Electric Flux through the Gaussian Surface.

    I am bit confused over there. If we have Gaussian Surface which is a cube placed in a non-uniform linear electric charge (by an infinite sheet for instance, and the Electric Field is parallel (anti-parallel) to the area vector of one of the faces (and the opposite face) ) the Flux through the two opposite faces of the cube does not cancel out. Is this true? What does the Gauss's Law actually states (along with conditions)?
    Last edited: Dec 9, 2014
  2. jcsd
  3. Dec 9, 2014 #2
    Gauss' Law for electrical flux is concerned with total flux through the whole closed surface, not through the opposite faces/parts of the surface.
  4. Dec 9, 2014 #3
    Yes of course not. I forgot to mention the direction of the Electric Field relative to the Cube. Please see the edited question.
  5. Dec 9, 2014 #4
    Net flux through the gaussian surface indicates wether it encloses a net charge or not. No net flux means the surface doesn't enclose any net charge.
    For example, if the gaussian surface is a cube placed in an external field, and if cube encloses no net charge, than flux through any two opposite pair of faces must cancel out regardless of direction of external field. BTW, your question is more appropriate for general physics subforum than here.
  6. Dec 9, 2014 #5
    zoki85 - But in the case of non-uniform electric field it does not cancel out the flux of opposite faces, does it?

    Can I somehow shift this question to the General Physics section?
  7. Dec 9, 2014 #6
    In general case of nonuniform fields, total net flux is 0 (if qin=0), but you have to integrate over all the faces of the gaussian surface.
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