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How much charge is enclosed by the box?

  1. Jan 31, 2013 #1
    1. A box with a side of length L=2.54 cm is positioned with one of its corners at the origin of a rectangular coordinate system. There is a uniform electric field
    E= 4 N/C(i) + 5 N/C(j) + 6 N/C(k).
    How much charge is enclosed by the box?


    2. I believe I should be using Gauss's Law (Epsilon=permittivity of free space)
    Ie= Q/Epsilon
    Ie= ExA
    Area=L^2



    3. I basically multiplied all sides with its proper vector x area. The area for all sides seems to be the same since all L=2.54cm=.0254m.
    Area came out to be 6.4516X10^-4 m^2.
    I came out with a charge=Q=0
    Not quite sure if I am calculating this correctly; please help.
     
  2. jcsd
  3. Jan 31, 2013 #2

    haruspex

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    Since you have not shown all your steps, it's impossible to say whether you are doing it correctly. Since the field is uniform, there is certainly no enclosed charge.
     
  4. Jan 31, 2013 #3
    Yes. That is correct to say that there is no charge since the electric field is uniform.
    Area was the same all around so the electric flux canceled each other out in the i, j, z directions. Q=0.
    Thanks for your help.
     
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