1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    User Avatar
    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
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook