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Charge inside parallelepiped

  1. Mar 12, 2015 #1
    1. The problem statement, all variables and given/known data
    The electric field ##\vec E_1 ## one face of a parallelepiped is uniform over the entire face and is directed out of the face. At the opposite face, the electric field ##\vec E_2 ## is also uniform over the entire face and is directed into that face. The two faces in question are inclined at 30.0° from the horizontal, while ##\vec E_1 ## and ##\vec E_2 ## are both horizontal; ##\vec E_1 ## has a magnitude of ##2.40*10^4 N/C## , and ##\vec E_2 ## has a magnitude of ##7.10*10^4 N/C##.

    Assuming that no other electric field lines cross the surfaces of the parallelepiped, determine the net charge contained within.

    YF-22-35.jpg

    2. Relevant equations
    ##\oint \vec E \cdot d \vec A = {\frac{Q}{\epsilon_0}}##
    ## A = lw ##

    3. The attempt at a solution

    ##\oint \vec E \cdot d \vec A = {\frac{Q}{\epsilon_0}}##
    ##EA = {\frac{Q}{\epsilon_0}}##
    ##Q = EA\epsilon_0##
    ## 2.4*10^4 - 7.1*10^4 = -4.7*10^4##
    ##Q = (-4.7*10^4)(6*10^{-2})(5*10^{-2})(sin(30°))(8.85*10^{-12})##
    ##Q = -6.24*10^{-10} C ##


    I already figured out the solution, but the only issue I have is the with the angle. I was convinced that I need to use cosine but after I got the answer wrong, I was told that I need to be using the sine function. I thought that since the E fields are moving in the horizontal direction, I would need to use cosine. What is the reason for using sine in this situation? Much appreciated!
     
  2. jcsd
  3. Mar 12, 2015 #2

    SammyS

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    What is the angle made by the electric field and the normal to the surface?
     
  4. Mar 12, 2015 #3

    TSny

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    The dot product of two vectors involves the cosine of the angle between the two vectors. So when considering ##\vec{E_1}\cdot\vec{dA}##, you need to find the angle between ##\vec{E_1}## and ##\vec{dA}##.
     
  5. Mar 12, 2015 #4
    Okay this is my interpretation...the angle made between ## \vec {E_1} ## and ## \vec {dA} ## is 30°, as given by the problem. Now that I made this picture, I think I can see it. But just to make sure, sin(30) here is ##{\frac{E_{1y}}{E_{1x}}}##

    So from there, I shouldn't have to take into account the angle formed by ##\vec {E_2}## and ##\vec {dA}## because I already found the net charge and it's going through the front...does that make sense or is there another reason not to include that angle?
     

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  6. Mar 12, 2015 #5

    TSny

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    Check your textbook or notes on the definition of the direction of the area vector ##\vec{dA}##.

    Or, see http://www.kellerphysics.com/AP_C/APC_notes_on_Gauss's_law.pdf

    In the figure below, does the green vector or the blue vector point in the direction of ##\vec{dA}##?
     

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    Last edited: Mar 13, 2015
  7. Mar 13, 2015 #6
    The area vector is the vector that is perpendicular to the area, so the green vector is ##\vec {dA}##

    The rest of the equation makes sense, I think. I'm just having trouble putting together the electric field portion.
    A is the area of the face
    ##\epsilon_0## is constant
    E should be net field times angle
    My textbook is basically saying the same thing as the link you sent, but I'm trying to see why cosine is used in the text and sine is needed here. Sine is opposite over hypotenuse, so in the example with the parallelepiped wouldn't that just be ##{\frac{\vec {E_{1y}}}{\vec {dA}}}##? Why wouldn't I want to use adjacent over hypotenuse (cosine), which would be ##{\frac{\vec {E_{1}}}{\vec {dA}}}##?

    Edit: I added an image of what I'm talking about in case I'm violating ten different rules of trigonometry and geometry
     

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    Last edited: Mar 13, 2015
  8. Mar 13, 2015 #7

    TSny

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    What is the angle in degrees between the E vector and the A vector? How does the cosine of that angle compare to the sine of 30o?
     
  9. Mar 13, 2015 #8
    Ohh that angle is 60°, and cos(60)=sin(30). I just need to make sure I'm using the correct angle...thanks for pointing that out!
     
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