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Gauss' Law and electric flux

  1. Jan 22, 2007 #1
    1. The problem statement, all variables and given/known data
    A charge Q is located inside a rectangular box. The electric flux through each of the six surfaces of the box is: Φ1=+1500 Φ2=+2200 Φ3=+4600 Φ4=-1800 Φ5=-3500 Φ6=-5400.
    (unit: N x m^2/C)
    What is Q?

    2. Relevant equations

    ΦE =Q/ε

    3. The attempt at a solution
    Add up all the Φ's to get -2400, then use ε0Φ=Q

    (-2400)(8.85x10^-12) = -2.124 x 10^-8 C
    Last edited: Jan 22, 2007
  2. jcsd
  3. Jan 22, 2007 #2


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    What are you supposed to do? Find Q? ε is a constant usually denoted [tex] \epsilon_0[/tex].
  4. Jan 22, 2007 #3
    yes im supposed to find Q

    Hows this:
    Add up all the Φ's to get -2400, then use ε0Φ=Q

    (-2400)(8.85x10^-12) = -2.124 x 10^-8 C
    Last edited: Jan 22, 2007
  5. Jan 22, 2007 #4
    [tex]\epsilon[/tex] is the permittivity of the medium enclosed by the surface which, in this case is the rectangular box. If the medium is air than it is equal to the permittivity of free space denoted by [tex]\epsilon_0[/tex]. For more info about permittivity see http://en.wikipedia.org/wiki/Permittivity

    You should start by thiking what does Gauss' law state? What does each quantity in the formula of the law represent?

    EDIT: looks like you solved it
    Last edited: Jan 22, 2007
  6. Jan 22, 2007 #5
    is that right? i thought i was supposed to get zero?
  7. Jan 22, 2007 #6
    Why would you get zero?
  8. Jan 22, 2007 #7
    I was reading this, and it seems similar to my problem

    Last edited: Jan 22, 2007
  9. Jan 23, 2007 #8
    In this problem the total flux through the surface of the box is 0, so the electric charge inside it will be 0.

    In your problem the total flux is non-zero so the charge inside will be non-zero.

    This is exactly what Gauss' law is all about.
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