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Homework Help: Electric Flux through cube

  1. Apr 26, 2010 #1
    1. The problem statement, all variables and given/known data
    Consider a cube of sides L. Suppose that a non-uniform electric field is present and is given by E(x) = {a(x+L)^2}x - (ayL)y where a is a constant.

    a) determine the elctric flux through the face of the cube that lies in the xy plane at z = 0 (express answer in a & L

    b) what is the total flux through the cube

    c) what is the volume charge density within the cube?

    d) if L=2 and the total charge within the cube is 70.8pC, determine the value of a

    2. Relevant equations

    3. The attempt at a solution

    a) determine the electric flux throught face on the xy plane (z=0)

    first off for the electric field the bold x & y im assuming are the same as the vector directions i & j so E(x) = {a(x+L)^2}i - (ayL)j

    [tex]\oint[/tex] E dA where E is noted above and dA = kdxdy = k2dx

    therefore the flux =0 because the electric field does not have a k component

    b) the total flux through the cube

    so for this i believe i only need to find it throught the surfaces of the xz & yz planes and then double it.... dA = dydzi + dxdzj = 2dxi + 2dxj
    [tex]\oint[/tex] {a(x+L)^2}i - (ayL)j dot dydzi + dxdzj from 0 to L

    = [tex]\oint[/tex] 2a(x+L)^2 dx - 2ayLdx from 0 to L
    = 2aL(x+L)^2 - 2ayL^2 which i must double to inslde the two oposite faces

    Flux = 4aL[(x+L)^2 - yL]

    c) find the volume charge density

    p = dQ/dV so it seems like i have to diferentiate Q interms of V, but there is not V so im not sure im on the right path

    d) when I plug the values in I am still left with the x & y variables. are these suppose to remain as variables or am i missing something
  2. jcsd
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