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Electric Flux Question

  1. Jul 9, 2006 #1
    The non uniformity of the electric field in the following question is throwing me off. If the electric field were uniform I'd have no problem.

    I assume I would use the following equation to solve for each of the surfaces:
    [tex]\Phi = \int \vec{E} \cdot d \vec{A}[/tex]

    I'm having a difficult time picturing what's going on. How do I know what surfaces the E field is passing through given just a vector? As far as I can tell there are four surfaces that are not parallel to the E field direction so four surfaces should have a flux value, right? The answer to the question a says there's only two surfaces with flux values not equal to zero. The answer to b says there's a charge in the cube. If there's a charge in the cube why doesn't its E field pass through all surfaces of the box it's enclosed in?
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  3. Jul 9, 2006 #2


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    Hello greenmaze,

    Start with calculating the scalar product of [itex]\vec {E}(\vec r)[/itex] and the normal vectors [itex]\vec n[/itex] of the cube's faces.

    [tex]\Phi = \int \vec{E} \cdot d \vec{A}=\int \vec{E} \cdot \vec{n}\,dA[/tex]


    Last edited: Jul 9, 2006
  4. Jul 9, 2006 #3
    What does x and z stand, coordinate values ?
  5. Jul 9, 2006 #4
    By [itex]\vec {E}(\vec r)[/itex] do you mean this?

    The top side of the cube has the dimensions (.3, .3, 0) so [itex]\vec {E}(\vec r)[/itex] would be:
    [itex](-5.00 N/C\cdot m)(.3 m)\hat{i} + (0 N/C\cdot m)(.3 m)\hat{j} + (3.00 N/C\cdot m)(0 m)\hat{k} = (-1.5 N/C) \hat{i} + 0 \hat{j} + 0 \hat{k}[/itex]

    Assuming that's correct, what would the vector normal to the top surface of the cube be? Is this a vector of magnitude [itex](.3 m)^2[/itex] in the positive z direction (i.e. [itex](.09 m^2)\hat{k}[/itex])?

    If both of those are true then the dot product would be [itex]0[/itex] since [itex](-1.5 N/C)(0) + (0)(0) + (0)(.09 m^2) = 0[/itex], which the answer doesn't agree with.

    Where am I going wrong?
    Last edited: Jul 9, 2006
  6. Jul 9, 2006 #5


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    You've got to find the flux over each face, by integrating. Remember that your Electric field is a function of x ,z . What you've got to do, is find the infentisimal area [tex]d\vec{A}[/tex] on each face of the cube, then find the dot product [tex]\vec{E}.d\vec{A}[/tex] on each of the six faces and integrate.
    Last edited: Jul 9, 2006
  7. Jul 9, 2006 #6


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    No, you are going about it the wrong way, I am sorry to say. First, you are not writing the correct normal vector; it must be a vector pointing normal from the surface, outward from the cube. You are using a vector with both an x and a y component, that's a vector which is tilted at 45 degrees...it is NOT perpendicular to any of the faces of the cube! Second, you are not taking into accound that the electric field vector is not constant.

    I wonder, has your prof done any example? I could not imagine giving a question like this to students without showing a few examples!

    Anyway, Siddharth explained how to do it but I will add a few details.

    Using a y axis pointing up, an x axis pointing to the right and the z axis pointing out of the page, the top surface has an infinitesimal area vector
    [tex] d{\vec A} = dx dz {\hat j} [/tex]
    and on that surface, y is fixed at the constant value of 0.3.
    To calculate the flux, you dot this with the electric field and integrate over x and z (both from 0 to 0.3). Again, y is not a variable on that surface, it is fixed at the value of 0.3 m.

    Does this make sense? If this is clear, you should be able to do all six faces (for example, the bottom face has [itex] d{\vec A} = -dx dz {\hat j} [/itex] and on that surface z =0)
    . It will be clear that two faces will give zero before even doing any integral.

  8. Jul 9, 2006 #7
    No, my prof hasn't done an example. Probably because there isn't enough time in a 5 week summer session. There are also no examples in my text dealing with non-uniform fields, which doesn't help.

    Thanks for the help. I think I can handle this problem from here.
  9. Jul 10, 2006 #8


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    Wow... then I feel for you:frown:
    This is a *very* difficult problem to do if one has never done any example! I hope that the two surfaces element Igave you are clear enough for you to see how things work. After you have the surface element, then you dot it with the E field and then you integrate over the surface. (actually, for your specific problem, things are quite simple because only 2 surfaces contribute and you end up not having to do any integral in the end! See below)

    For example, for the top surface, [itex] d {\vec A} = dx dz {\hat j} [/itex] as I mentioned before. Therefore th eflus through the top surface is
    [tex] \int_0^{0.3} \int_0^{0.3} E_y dx dz [/tex]
    all this with the y value set equal to 0.3 (so y is not a variable).

    But in your example, the E field has no y component so this is zero!!
    For the same reason (no y component of the E field), the flux is zero through the bottom surface. This all makes sense because the electric field is always "horizontal" everywhere (i.e. it has no y component anywhere) therefore it does not "cross" the top and bottom faces of the cube so no flux through those two surfaces.

    However, there *will be* some nonzero flux through the other faces.

    Just as another example, through the "right" face, [itex] d{\vec A} =dy dz {\vec j} [/itex] and the flux through that face is
    [tex] \int_0^{0.3} \int_0^{0.3} dy dz E_x [/tex]
    all this with x fixed at the value 0.3.

    In *your* specific example, the integral is trivial because E_x will be a constant which can be pulled out of the integral and you are left with E_x evaluated at x=0.3 times the area of the surface!!!
    So your problem is actually quite simple and that's because the x component of the E field depends only on x (not on y and z) and the z component of the E field depends only on z (and not on x or y).

    Actually, I just noticed that the flux will be nonzero only on *two* of the 6 faces (not on 4) in your problem!

    If you have any question, let us know.

  10. Sep 23, 2006 #9
    Hey, I found this problem and it was exactly like mine except for different numbers. I was wondering in how would you go to solve part B? I thought it would be q = ( flux * L^3 ) / E_0 but it wasn't. Could anyone possibly help me with this?
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