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Electric flux - Area vector question

  1. Jul 18, 2012 #1
    This is my first post here.

    I have a problem dealing with Electric flux.

    According to Gauss's Law, to find electric flux we use equation:

    ψ=∫D dot dA, both D(Electric flux density) and dA(super small area normal to the surface) are vector quantity.

    Now this is what I confused, how can "dA" be a vector quantity since it's an area.

    I may misunderstand something.

    Thank you!
  2. jcsd
  3. Jul 18, 2012 #2

    Philip Wood

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    Gold Member

    The electric flux [itex]d\Phi[/itex] through an area dA is defined as [tex]d\Phi=D\normalsize{d}A cos \phi[/tex].
    [itex]\phi[/itex] is the angle between the normal to the area and [itex]\textbf{D}[/itex].
    [itex]D cos \phi[/itex] is the component of D at right angles to the area.

    We can write [itex]d\Phi=D\normalsize{d}A cos \phi[/itex] more neatly, as
    if we define a vector [itex]d\textbf{A}[/itex] such that [itex]d\textbf{A}=\textbf{n}dA[/itex].

    Here, [itex]\textbf{n}[/itex] is a unit vector normal to the area. If the area is part of a closed surface, we choose the normal pointing outwards from the enclosed volume.

    Hope this helps. Congratulations on your first post. It is very clear.
    Last edited: Jul 18, 2012
  4. Jul 18, 2012 #3
    http://www.technology2skill.com/science_mathematics/vector_analysis/vector_picture/curve_surface_integral_first.png [Broken]

    This might be a helpful picture: Here k is a flux vector, and n is the unit vector normal to the surface.

    The reason why dA is a vector quantity is the amount of electric flux moving through a given Gaussian surface depends not only on the magnitude of the electric flux, but also its direction. For example, if the E-field is flowing directly perpendicular to the gaussian surface, there can't be any flow through the surface, and the flux will be zero. (You can see why the dot product/cos is used here!)
    Last edited by a moderator: May 6, 2017
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