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Calculate Flux in a closed Triangle

  1. Oct 12, 2009 #1
    1. The problem statement, all variables and given/known data

    Consider a closed triangular box resting within a horizontal electric field of magnitude E = 7.80 & 104 N/C as shown in Figure P24.4. Calculate the electric flux through (a) the vertical rectangular surface, (b) the slanted surface, and (c) the entire surface of the box.


    2. Relevant equations

    Flux = EAcos( )

    3. The attempt at a solution

    for a) I did

    (7.8E4)(.1x.3)cos(0) = 2.34kN

    The answer should be negative, why?
    Would it be cos(180) instead? What is the reason for this?

    for b), the slanted plane, the answer is 2.34kN, but why would it be? I did (7.8E4)(.1x.3)cos(0), which gives the right answer, but why is it 0? not 60 or something?

    c) I am assuming because a and b cancel eachother when added together.

    A little input would be nice, thanks :)

  2. jcsd
  3. Oct 12, 2009 #2
    a) The surface normal and the electric field point in opposite directions.

    b) The area of the slanted surface is not [tex]0.1 \times 0.3[/tex] but [tex]\frac{0.1}{\cos 60} \times 0.3[/tex]

    [tex]\Phi_E=(7.8\times 10^4) \times \frac{0.1}{\cos 60} \times 0.3 \times \cos 60=(7.8\times 10^4) \times \0.1\times 0.3[/tex]
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