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Integral of E dot dA - conceptual

  1. Oct 5, 2009 #1
    1. The problem statement, all variables and given/known data
    A = 2i + 3j
    E = 4i

    determine the integral of E dot dA

    2. Relevant equations
    Integral calculus, vectors


    3. The attempt at a solution
    I don't understand why one could do this. The integral is of E and dA, not E and A. How can I use A to determine dA? Do I take its derivative? Then I would only get dA = 0.


    Also: This is part of an electric flux problem. Is this a basic of multivariate calculus? I'm indecisive on whether or not I should get a MV calc textbook - I don't have one yet, but I don't want basic questions like this holding me back either. I'm studying from Halliday's Fundamentals of Physics which apparently isn't a calculus heavy text.
     
    Last edited: Oct 5, 2009
  2. jcsd
  3. Oct 5, 2009 #2

    lanedance

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    Homework Helper

    the question isn't very celar, but here's my 2 cents...

    E is the electric field vector

    Generally A represent some surface or area A. dA is vector in the direction normal to surface. If n is a unit vector in that direction, dA is a scalar infintesimal area, then

    [tex] \texbf{dA} = \textbf{n}dA [/tex]
    with
    [tex] A = \int dA [/tex]

    For a vector field E, the flux is given by
    [tex] \Phi = \int \textbf{E} \bullet \textbf{dA} [/tex]

    physically the dot product means if the vector is parallel to the surface its contribution is zero, or maximum when it is perpindiuclar to the surface (passing through it). The integral just sums all the contributions across the whole surface.

    so is there anymore to make the question clearer?
    what area do you integrate over?
    A is given as a vector - does this represent a normal to a plane or something?
     
  4. Oct 6, 2009 #3

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    You are right. The problem as stated makes no sense. I suggest you go back and reread the problem. It may be that you have overlooked something. "dA" should be a "vector differential of area" of some surface, not just a vector. Are you not given a surface to integrate over?
     
  5. Oct 6, 2009 #4
    I'm sorry. dA is in fact an area vector perpendicular to a surface. However, the surface is definitely not stated. The answer is given as 8 (flux units). Does it make any mathematical sense to do this?

    [tex] \Phi = \int \vec{E} \cdot \vec{dA} [/tex]

    [tex]
    \Phi = \vec{E} \cdot \int \vec{dA}
    [/tex]

    [tex]
    \Phi = \vec{E} \cdot \vec{A}
    [/tex]


    Then
    Phi = (4i) dot (2i + 3j)
    = 8

    No clue if this makes any sense or not but...?
     
    Last edited: Oct 6, 2009
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