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Homework Help: What is the electric flux?

  1. Oct 1, 2008 #1


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    1. The problem statement, all variables and given/known data
    A 2.6-uC charge is at the center of a cube 7.5cm on each side. What is the electric flux through one face of the cube?

    q_enclosed = 2.6*10^-6 C
    a = .075m

    2. Relevant equations
    flux = E*A = integral(E*dA*cos(theta)) = E*integral(dA)
    flux = q/(e_0)
    surface area of a cube: 6*a^2

    3. The attempt at a solution
    flux = E*int(dA) = E*6(.075)^2 = .03375E

    .03375E = (2.6*10^-6)/(e_0) => E = 6.8*10^-16 N/C

    So the flux = q/(e_0).

    ... you know what. I just figured it out.

    flux = q/(e_0) => (2.6*10-3C) / (e_0) = 294053 Nm^2/C

    (1/6)(294053 Nm^2/C) = 4.9*10^4 Nm^2/C

    So the flux through one side of the cube (any cube!) with the enclosed charge of 2.6-uC is 294053 Nm^2/C.
    No integral is necessary.

    edit: I was going to not post this but maybe it will help someone out.
  2. jcsd
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