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Homework Help: Determine the electric flux through each of the sides

  1. May 21, 2007 #1
    A 2.10- mC charge is placed at the center of a cube of sides 6.40 cm. Determine the electric flux through each of the sides.

    ok so electric flux=E(A)

    so i figure that the electic flux will be the same for all six sides of the cube so i am just calculating the flux on one side then X6.

    so that would be =(2.10x10^-6)(.064)(.064)=8.60x10^-9...for one side

    and that would be 5.16x10^-8 N m^2/C for all 6 sides

    but this isn't right...please tell me what i am doing wrong
     
  2. jcsd
  3. May 21, 2007 #2

    hage567

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    So what is the charge? It looks like you wrote 2.10 mC, in the question, and then used 2.10x10^-6 C in the calculation.

    You never actually calculated the electric field due to the point charge. The E in EA is electric field, in N/C. Check your units, they don't work out to the units of flux the way you have it.
     
  4. May 21, 2007 #3
    ohh thats right...ok so now if i calcuate what E is

    E=Kq/r^2

    now r is at the the center of the cube so r=.064/2

    E=(8.99x10^9)(2.1x10^-6)/.032^2=1.84x10^7

    A=.064*.064=.004096

    now 1/6flux=(1.84x10^7)(.004096)

    flux=4.53x10^5....but that still isn't the right answer

    help me please
     
  5. May 21, 2007 #4

    hage567

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    [tex]\Phi=\oint{\bf E} \cdot{d}{\bf A}=\frac{q}{\epsilon_o}[/tex]

    Look closely at this. What does it say?
     
  6. May 21, 2007 #5
    that means that the flux= the charge on the inside of the soild/ the constant e which is 8.85 x10^-12
     
  7. May 21, 2007 #6

    hage567

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    Yes, so the net electric flux through any closed surface is equal to the net charge inside the surface divided by [tex]\epsilon_o[/tex].

    How would you go about solving the problem knowing this? This makes it MUCH easier!
     
  8. May 21, 2007 #7
    well the net charge inside is just 2.1x10^-6 C/e is that right?
     
  9. May 21, 2007 #8

    hage567

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    That's right. As long as you're sure it's 2.10 [tex]\mu C[/tex], and not 2.10 mC.
     
  10. May 21, 2007 #9
    right it is the first value...i don't have the notation for that in my keypad....


    but the when i do that operation =2.37x10^5 N m^2/C which is wrong
     
  11. May 21, 2007 #10

    hage567

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    OK, but that is for the flux through the entire surface.
    I think you want the flux through one side, so divide by six.
     
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