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B How to find the appropriate area in Gauss' law

  1. Feb 1, 2017 #1
    Knowing that Gauss' law states that the closed integral of e * dA = q(enclosed)/e naut, how would you find exactly what A is in any given problem?

    I know it varies from situation to situation depending on the geometry of the charge. For instance, I know that for an infinite wire/line of charge, the area will be 2*pi*r*l.

    While I understand that this is the area of the lateral sides of the cylinder that serves as the Gaussian surface in this particular example, I don't understand why the ends don't count. I've heard explanations such as the ends are parallel and that the electric field would therefore never go through them. This explanation doesn't make sense to me, however, because (and correct me if I'm mistaken) the electric field given off of a charge is multi-directional.

    Since the electric field is multi-directional, wouldn't at least SOME of the electric field go through the ends? From my understand the "they're perpendicular so the electric field doesn't go through them" explanation would only make sense if the electric field were strictly 90% to the ends, otherwise some of the field should go through.
     
    Last edited by a moderator: Feb 1, 2017
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  3. Feb 1, 2017 #2

    stevendaryl

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    It's a symmetry argument. For definiteness, let's imagine an infinite charged wire sitting on the x-axis, and orient the x-axis so that it runs left-to-right. Take a point [itex]\mathcal{P}[/itex] not on that wire. Now, divide the wire into two halves: the half that is to the left of the wire, and the half that is to the right of the wire. At point[itex]\mathcal{P}[/itex], there will be an electric field [itex]\vec{E}_L[/itex] due to the left half, and a field [itex]\vec{E}_R[/itex] due to the right half. The total electric field at point [itex]\mathcal{P}[/itex] is the sum of these two. The point is that the x-components cancel perfectly: [itex]E_L^x = - E_R^x[/itex]. So for the total field, the only nonzero component is radially away from the wire.
     
  4. Feb 2, 2017 #3
    Gauss's law does not "require" or dictate any particular shape or type of area. The law says that whatever closed surface you choose, the flux of the electric field over the surface is proportional to the enclosed charge. So in the particular case of an infinite line charge that you discussed, Gauss's law is true even if you take an irregular shaped surface which encloses a part of the line charge.
    If you want to get a "useful" result from Gauss's law, then you should choose a surface which conforms to the symmetry of the charge distribution. For a point charge, you choose a sphere with the charge at the center. For a line charge, you choose a cylindrical surface with the line charge along the axis of the cylinder (see the symmetry argument by stevendaryl above). Such a choice helps you to calculate the electric field, given the charge distribution, or the charge, given the electric field.
    If the charge distribution itself is not symmetric, but is irregularly shaped, Gauss's law is still valid, but is not useful for calculating anything.
     
  5. Feb 2, 2017 #4
    I've seen the infinite wire example before, but it's dangerous because you kinda break Gauss' Law (an infinite wire can't have an enclosed, finite area around it), but then sorta "unbreaks" it with those symmetry considerations.
     
  6. Feb 2, 2017 #5
    There is no danger. The wire is infinite, which is what gives the symmetry of the field. The Gaussian surface is a finite length cylinder enclosing a finite part of the infinite wire. You then apply Gauss's law to the finite cylindrical Gaussian surface, and the flux of the electric field is equal to the finite amount of charge enclosed by the cylinder. You will see the problem worked out in any introductory physics textbook.
     
  7. Feb 3, 2017 #6
    Oh, I see. The enclosing area is finite because the symmetry lets you do that, as no field lines will cross the caps. Never mind my comment then.
     
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