1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Electro Box

  1. Jan 8, 2008 #1
    [SOLVED] Electro Box

    1. The problem statement, all variables and given/known data:
    A cubic box of side a = .290 m is placed so that its edges are parallel to the coordinate axes. There is NO net electric charge inside the box, but the space in and around the box is filled with a nonuniform electric field of the following form: E(x,y,z) = Kz j +Ky k, where K=4.10 N/(C*m) is a constant.


    2. Relevant equations:
    [tex]\oint[/tex][tex]E[/tex][tex]\bullet[/tex][tex]dA[/tex] = Q/E_naught



    3. The attempt at a solution
    I'm just inquiring as to where to begin this problem. I've only taken through a high school equivalent of first semester calculus, so I have no idea how this equation works. If anyone could point me in the right direction, I'd be very greatful!
     
  2. jcsd
  3. Jan 8, 2008 #2
    You...didn't ask a question? I'm assuming it's to find flux since that's the equation you gave

    You're given a vector field in the form of E=K*zj+K*yk, j and k are the y and z unit vectors, respectively, and K. This is just given for the problem, it has no other fundamental significance. It's a simple vector field, so if you're unfamiliar, google or wiki them and try to figure out what this one looks like

    The equation you have is Gauss' Law, that integral on the left that may look funky to you is a surface integral, the loop in the middle represents that it's a closed surface(which a cube is!)

    http://en.wikipedia.org/wiki/Gauss's_law

    That first line of math stuff is just what you've typed there, note, the derivation and understanding are useful(why flux through a closed surface is what it is)

    EDIT: This is a little weird without knowing the question >_> I guess it's show that with that electric field the charge inside the box is 0? It would be weird to say "here's a non-uniform electric field" and "there's no charge inside the box" and then have one of the fundamental laws of nature point out there must be a charge inside the box. So I guess you'll do that work and find Q=0 despite E=/=0. I think. Been a while >_>

    http://en.wikipedia.org/wiki/Gaussian_surface

    that'll give you some help on solving it, note that you'll have six integrals for the six sides, and the integral of 1*dA over a side will be just the area of the side. I don't think it'll come up but that maybe illustrates what is meant by dA, it's the infinitismal area "vector", which, btw, points out of the surface. So the top side's dA is in the positive z direction, the bottom side's is in the negative
     
    Last edited: Jan 8, 2008
  4. Jan 8, 2008 #3
    Oh man, sorry my bad about not posting the question. Our capa problems are created...oddly. I thought I grabbed it, but I suppose not >_< The actual question it asks is:
    What is the electric flux through the top face of the box? (The top face of the box is the face where z = a. Remember that we define positive flux pointing out of the box.)

    Anyways, I'll look into the resources you provided. Thank you very much!
     
  5. Jan 8, 2008 #4
    Do you know the answer? The way I figure, and it's been a while since I've done these, since there's no net charge in the box, there's no net flux in or out of the BOX, but the individual faces are different.

    I believe you'd break down into the six integrals for each face, and each integral would represent the flux through that particular face? <--that's where I'm thinking I messed up if I did

    so you'd say da=da*z(oh hey I'm gonna use bolded x, y, z instead of i, j, k, I'm newschool like that)

    the dot product of E and da will then result in just the z component, ky(giggle) left times da. For the top surface da is dxdy and it's now a double integral, the dx part just gives you a, and the dy part, since we have y in the integrand, leads to a net result of a^3*k/2

    Lemme know if that's clearly way too much math for it to be right(it's an easy double integral but if you've never had one I probably did something wrong)or if you know the answer is wrong, so I can delete this before exposing my mistakes! :)
     
  6. Jan 10, 2008 #5
    Sorry about not getting back to you sooner - our school's internet was down yesterday.

    Anyways, I don't know the answer, and as of right now, and I've no idea if that math is correct or not. I think my problem lies in not understanding line integrals very well, as we've only done basic intro calc stuff. Where might be a good place to look for figuring those out?
     
  7. Jan 10, 2008 #6
    I think the math is fine, I'm just not sure that it's actually the flux or anything meaningful that I found

    Also, it's a surface integral, not a line integral, but similar idea. If it's just physics 2 they usually give you absurdly simple ones since they're in general not necessarily very easy, and often hard. Things like where the surface integral just ends up being the surface area of like a sphere
     
  8. Jan 14, 2008 #7
    Ahhhhhhh, ok. That makes sense. It seemed really strange that they would give us math like that. Thanks for your insights!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Electro Box
Loading...