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How much charge inside a Gaussian surface?

  1. May 1, 2015 #1
    1. The problem statement, all variables and given/known data
    The electric field has been measured to be horizontal and to the right everywhere on the closed box shown in the figure. All over the left side of the box E1 = 80 V/m, and all over the right, slanting, side of the box E2 = 400 V/m. On the top the average field is E3 = 260 V/m, on the front and back the average field is E4 = 200 V/m, and on the bottom the average field is E5 = 230 V/m.

    2. Relevant equations
    Flux = |E|*|A|*cos (theta)
    ε0 = 8.85 ✕ 10-12 C2/N·m2
    3. The attempt at a solution
    The flux on the sides of the box = 0 because the electric field is perpendicular to the area vector. The flux on the left side of the box is (80)(.002)(cos 180) = -0.16 The flux on the right side of the box is (400)(.006)(cos (inverse sine of 4/12)) = 2.26
    Then net flux is -0.16 + 2.26 = 2.1
    2.1 = Q / (8.85x10^-12), Q = 1.86x10^-11

    I can't figure out where I went wrong with this!
     
  2. jcsd
  3. May 1, 2015 #2

    ehild

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    There is no figure.
     
  4. May 3, 2015 #3
    Here is the figure!
     

    Attached Files:

  5. May 3, 2015 #4
    You need the sine of the inverse sine, so 1/3.
     
  6. May 3, 2015 #5

    Zondrina

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    The electric field is uniform, so I believe you have the correct equation.

    Notice that ##\vec E_3, \vec E_4##, and ##\vec E_5## cause no electric flux because each area vector ##d \vec A_{3, 4, 5}## is perpendicular to the electric field.

    Only ##\vec E_1## and ##\vec E_2## are going to cause any flux.

    I believe you have calculated the electric flux caused by ##\vec E_1## properly.

    As for the electric flux caused by ##\vec E_2##, you need to separate the area vector for that side of the surface into it's ##x## and ##y## components ##d \vec A_x## and ##d \vec A_y##. Only the ##x## component should allow you to calculate a flux because the angle between it and ##\vec E_2## is zero.
     
  7. May 3, 2015 #6
    I think I may have found my mistake: wouldn't Flux2 = (.12) (.05) (400) cos (90 - (inverse sine (.04/.12)) = .01424 ? Then net flux would be -0.1458?
     
  8. May 3, 2015 #7
    Ok so I've checked over this several times, and I plugged it into my calculator wrong, so Flux2 = 0.8. However, it is still wrong! Does anyone know if there's some trick to it that I'm missing?
     
  9. May 4, 2015 #8

    Zondrina

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    I believe it is because the electric flied lines through that face are not uniform. You can't apply ##\Phi_{\vec E_2} = E_2 A \text{cos}(\theta)## directly because finding the angle between ##d \vec A = \vec n \space dA## and ##\vec E_2## is difficult.
     
  10. May 4, 2015 #9

    BvU

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    I can't see what's wrong, you don't show your calculation nor the result.

    It is given that the field is to the right and has the same value everywhere on the right side of the thing. I don't see why the 0.8 of love is incorrect.
     
  11. May 4, 2015 #10
    The answer is -0.16+400*0.006*1/3=0.64.
    See my previous post.
     
  12. May 4, 2015 #11

    BvU

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    Yes, the -0.16 from love's post #1 and the 0.8 from love's post #7 combined. So if the answer is wrong it must lie in the conversion from net flux to charge.
     
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