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Gaussian surfaces

  1. Aug 10, 2014 #1
    why is the electric field zero , if the closed gaussian surface doesnt enclose any charges.
    but,, if the charges are present at a distance outside the conductor??

    and why isnt the electric field zero inside the gaussian surface, if there are charges distributed on the surface of the gaussian surface..

    assume that our gaussian surface is an enclosed sphere.....
  2. jcsd
  3. Aug 10, 2014 #2


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    Your questions seem very confused, you mention a conductor...what conductor?

    I'm going to venture a guess that you're confused by Gauss's law. Gauss's law says that the surface integral of the electric field over a closed surface is equal to the charge enclosed by this closed surface.

    If there is no charge enclosed by the Gaussian surface (which doesn't have to be a physical surface like a conductor, but just any closed surface you want to draw) then the surface integral of the electric field is 0. It doesn't say that the electric field is 0, it says that the integral of the electric field over the Gaussian surface is 0. In the language of field lines, if there are no enclosed charges, then every field line that enters the Gaussian surface must exit it as well, so that the total integral is 0.
  4. Aug 10, 2014 #3
    can u pls explain the Gauss law????????? i am definately confused by it..
  5. Aug 10, 2014 #4


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    Gauss's law states, quite simply, that the surface integral over a closed surface of the electric field in space (let's just treat the vacuum for now, we can modify it if you're in a medium) equals the enclosed charge inside the surface. In mathematical language the law is (using SI units):

    $$\oint_S \vec{E}(x,y,z) \cdot d\vec{A} = \frac{Q_{\text{enc}}}{\varepsilon_0}$$

    Let's break this expression down. First you start off knowing that there is some electric field that permeates space (the electric field might turn out to be 0, if there are no charges anywhere). It is a vector field (it associates a vector with each point in space, even if that vector might be 0).

    Then you pick ANY closed surface (the surface must be closed, so you can't pick a plane or something, but you can pick a sphere, or a box, or a cylinder) and you take the integral of the electric field over this surface. When you are taking this integral, this is a surface integral, so you have to first take the dot product of the electric field vector with the surface element vector (which is a vector that is normal to the surface itself). This dot product tells you by how much the electric field "pierces" your surface at that point. You add up all these contributions from all over your enclosed surface, and you will obtain a single number (with units of V*m). Gauss's law says that this number is the same as the number you get if you take the charge enclosed by the surface you drew and divide it by the permittivity of free space (an artifact of the SI system of units, but let's not worry about that right now).

    Now, Gauss's law is completely general. You can pick any shaped surface you want (as long as it's closed) and any charge distribution you want and Gauss's law will still hold. However, unless your problem has some good symmetries involved (e.g. spherical symmetry, or cylindrical symmetry), it is very difficult, in practice, to actually perform that surface integral! You want to pick your Gaussian surfaces to match the symmetry of the given charge distribution. So if the charge is a point charge, choose a spherical Gaussian surface. If the charge is an infinite plane, then choose a pillbox.
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