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How to choose Gaussian surface

  1. May 31, 2013 #1
    1 Why do we choose a spherical surface as gaussian surface for a point charge to calculate electric field?

    In my view, the reason may be
    i. If we take the point charge at centre, each point of spherical surface will be at same distance from the point charge and thus is equivalent.Thus electric field (perpendicular to the surface ) at every point of the surface will be equivalent .

    If we take other surface like cube , the distance between the point charge and every point of cube will be different.Then the calculation becomes hard.

    2 Why do we choose a cylinderical surface for a infitely long char?ged sheet , why not other closed surface e.g. cube?
     
  2. jcsd
  3. May 31, 2013 #2

    vanhees71

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    This is all about symmetry. You can calculate the electric field for such highly symmetric situations with help of the integral form of Gauß's Law, because you can guess, how it should look like from the symmetry of the problem.

    For a point charge, everything is radially symmetric, and thus the electric field must point radially out from the charge, i.e., you must have
    [tex]\vec{E}(\vec{x})=E_r(r) \frac{\vec{x}}{r},[/tex]
    where [itex]r=|\vec{x}|[/itex]. Then you choose a sphere around the charge, because this gives you the missing component [itex]E_r(r)[/itex]. According to Gauß's Law you have (in Heaviside-Lorentz units)
    [tex]Q=\int_{S_r} \mathrm{d}^2 \vec{F} \cdot \vec{E}=4 \pi r^2 E_r(r)\; \Rightarrow \; E_r(r)=\frac{Q}{4 \pi r^2}.[/tex]
    thus you get the Coulomb field as expected
    [tex]\vec{E}=\frac{Q}{4 \pi r^3} \vec{x}.[/tex]

    For the infinite line charge the same idea holds. Due to symmetry the field must point radially out from the line, and you can use a cylinder as the surface in Gauß's Law.

    For more complicated situations, of course, you cannot guess the field's direction so easily anymore, and then you must apply the local form of the laws and integrate the corresponding differential equations.
     
  4. May 31, 2013 #3

    Doc Al

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    Staff: Mentor

    Just about any shape would work as long as the surfaces were parallel and perpendicular to the field. A cube would work fine, as long you oriented it correctly. (A cube at some odd angle to the sheet wouldn't work.)
     
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