Gaussian surface without any charge inside it

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Solution 1

No it is not necessarily 0. You can think about Gauss Law this way: It relates the total flux of the eletrical field to the net charge inside the surface, it is a direct relation between eletrical flux and charge inside. The flux is necessary 0, not the eletric field. Picture a point charge and a gaussian surface beside it ( but not with it inside ), the field is clearly not 0 at all points of the surface but the flux is.

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Solution 2

Consider the field around a point charge $$ \mathbf{E} = \frac{q}{4\pi \epsilon_0} \frac{\hat{\mathbf{r}}}{r^2}\;. $$ This is not zero anywhere (except at infinity). Now consider any closed surface which does not enclose the charge. The total charge enclosed is therefore 0 but at no point on the surface is the field 0. However the net flux entering and leaving the surface is 0.

Solution 3

No, it doesn't --- it means that there's an equal flux of electric fields leaving the surface and entering the surface. For an example, try constructing some Gaussian surfaces around an electric dipole.

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Updated on August 18, 2022

Comments

  • user164780
    user164780 about 1 year

    Suppose a Gaussian surface has no net charge inside it. Does it mean that the electric field $E$ is necessarily zero at all points on the surface? And is the converse also true? Can this be shown mathematically?

  • user164780
    user164780 over 6 years
    How will I prove this mathematically?
  • user164780
    user164780 over 6 years
    And it seems that you did not get my question.I want to know whether I can prove mathematically that the electric field intensity is not necessarily 0 on the Gaussian surface even if there is no charge enclosed by the surface.
  • rob
    rob over 6 years
    Look at any diagram of the electric field around a dipole. Draw a Gaussian surface someplace where the field isn't zero. Compute the flux integral --- you get zero if the surface encloses both charges and nonzero if you only enclose one charge.
  • user164780
    user164780 over 6 years
    I am not trying to bring "electric dipole " into consideration.
  • user164780
    user164780 over 6 years
    And in your case charge(s) is enclosed......but in my question I said that the Gaussian surface should not enclose any charge .
  • rob
    rob over 6 years
    Your question (v1) says "no net charge." I was suggesting taking equal-but-opposite charges and separating them by some nonzero distance, but I used a distracting technical name. But if you're integrating around a dipole field, you'll also get zero net flux if your Gaussian surface encloses neither charge.
  • user164780
    user164780 over 6 years
    Quite true....my mistake.....yes we can consider an electric dipole
  • user164780
    user164780 over 6 years
    And one more thing ...what about the electric field on the Gaussian surface as specified in the question.
  • user164780
    user164780 over 6 years
    Why is it not zero?
  • By Symmetry
    By Symmetry over 6 years
    We have an explicit formula for the field at every point in space, and it is never 0.
  • user164780
    user164780 over 6 years
    I mean why is the field intensity not zero at this Gaussian surface which is not enclosing any charge?
  • By Symmetry
    By Symmetry over 6 years
    Because the field intensity is not zero anywhere... Gaussian surface included