Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Flux through Cube

  1. Apr 27, 2010 #1
    1. The problem statement, all variables and given/known data
    A cube has lengths L.

    E(x) = {a(x+L)2i - (ayL)j where a is constant

    Find the total Flux through the cube.

    2. Relevant equations

    3. The attempt at a solution

    i know the flux is 0 through the z = 0 and z = L planes but the other sides and the top & bottom i am unsure how to find.

    I am confused about the x,y, & L variables. It seems that x & y = L. When I integrae over the area am i doing a double integral for x & y from 0 to L
  2. jcsd
  3. Apr 27, 2010 #2


    User Avatar
    Science Advisor

    There is NO "L variable", L is a constant, not a variable.

    If you are integrating the flux over the top and bottom, yes, you are integrating x and y from 0 to L.
    In particular, [/itex]\vec{n}dS[/itex] for the top and bottom are [itex]\vec{k}dxdy[/itex] and [itex]-\vec{k}dxdy[/itex] respectively. Since your given function has no [itex]\vec{k}[/itex] component, yes, the flux through those is 0.

    Now consider the face x= 0, in the yz-plane. A unit (outward) normal is [itex]-\vec{j}[/itex]
    and [itex]\vec{n}dS= -\vec{j}dydz[/itex].

    [itex]E\cdot \vec{n}dS= -(x+ L) dydz[/itex] but we are in the face x= 0 so that is just [itex]-Ldy dz[/itex] and the flux is [itex]\int_{y=0}^L\int_{z=0}^L -L dy dz[/itex] which is just -L times the area of the face, [itex]-L^3[/itex]

    On the face x= L, everything is the same except that the outward normal is [itex]\vec{k}[/itex] rather than [itex]-\vec{k}[/itex] and x+ L is now 2L. The flux through that face is [itex]\int_{y= 0}^L \int_{z= 0}^L 2L dydz[/itex] which is 2L times the area of the face, [itex]2L(L^2)= 2L^3[/itex].

    On the face y= 0, in the xz-plane an (outward) normal is [itex]-\vec{j}[/itex] and [itex]\vec{n}dS=-\vec{j}dxdz[/itex].

    [itex]E\cdot\vec{n}dS= -a(0)L(-1) dxdz= 0[/itex] so the flux through that face is 0.

    Finally, on y= 1, the (outward) normal is [itex]\vec{j}[/itex] and [itex]\vec{n}dS= \vec{j}dxdz[/itex].

    [itex]E\cdot\vec{n}dS= a(L)(L) dxdz[/itex] so the flux through that face is [itex]\int_0^L\int_0^L aL^2 dxdz[/itex] which is [itex]aL^2[/itex] times the area of the face: [itex](aL^2)(L^2)= aL^4[/itex].

    The flux through the entire region is the sum of those.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook