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Homework Help: Tough Flux question

  1. Feb 20, 2008 #1
    [SOLVED] Tough Flux question

    1. The problem statement, all variables and given/known data
    A cubic cardboard box of side a = 0.250 m is placed so that its edges are parallel to the coordinate axes, as shown in the figure. 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= 3.20 N/C.m is a constant.

    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.)

    Hint:Since E is not constant, you will need to do an area integral for this problem. Remember the definition of electric flux. Only one component of E contributes to the flux out of the top of the box. In doing the integral to find the flux, it might help to imagine breaking the top surface into strips along which the z-component of the electric field is constant. If all else fails, ask for help.

    2. Relevant equations

    3. The attempt at a solution
    The Nonuniform part is what's tripping me up. I know how to do it with uniform electric field.

    I'll call the top face faceA, Flux=I. So I[tex]_{A}[/tex]=[tex]\int[/tex]E*dA= [tex]\int[/tex](3.20 N/C z [tex]\widehat{j}[/tex] + 3.20 N/C y[tex]\widehat{k}[/tex])* (dA [tex]\widehat{k}[/tex])

    So it says that for this face Z=the side length, but what do I do with y? and with the top face wouldnt only the K-hat vectors apply anyway?

    and once I get an asnwer will it be negative?

    anyway if someone could give me some insight, let me know if ive set it up right it would help a lot. Thanks in advance!
    Last edited by a moderator: Apr 23, 2017
  2. jcsd
  3. Feb 20, 2008 #2
    I'm also really confused by the z being with the j and the y being with the k.
  4. Feb 20, 2008 #3
    Flux is the surface integral of E dotted with the normal vector. So, find E dot n first.

    Then, I think, you have to do a simple double integral to integrate over all the points on the surface. (edit: or instead, since the integrand only depends on y, you only need to integrate over y and then just multiply by a. It's the same thing really.)

    Also, you should write entire expressions in latex tags, not switch to it only to put symbols in.
    Last edited: Feb 20, 2008
  5. Feb 20, 2008 #4
    yeah i don't really know how to use the tags yet. And I though flux was E dot A, where does the normal vector come into play? :S thanks
  6. Feb 20, 2008 #5
    Well, the dA in the integral you use is really a vector quantity. It points in the direction perpendicular to the surface. So, when you take the dot product of E and dA, the normal vector I was talking about comes into play.
  7. Feb 20, 2008 #6
    so the dot product would be: (0.8 j + 3.20 y k)(.25^2) times cos90

    am I starting it right?
  8. Feb 20, 2008 #7
    Nope, because that quantity is 0. Remember that the theta you use is between E and dA (which is perpendicular to the surface). They are parallel.

    Also, keep in mind that the dot product is a scalar quantity, so there is no more vector stuff in it. You need to multiply together the magnitudes of the two vectors, but I don't think that will work for this expression.

    The best way to find the dot product is to use the other formula for it, where you sum up the products of their components.
  9. Feb 20, 2008 #8
    do I sub anything in for z and y yet?

    i got 3.20z +0.0625y as my dot product without subbing anything in.

    Sorry i'm being a noob :S
  10. Feb 20, 2008 #9
    I don't know where you got those numbers from. It's an easy dot product:

    [tex]\vec{E} \cdot d \vec{A} = <0, Kz, Ky> \cdot <0, 0, dA>[/tex]

    Note that the dA's are different.

    Then to integrate, have you done double integrals yet?
  11. Feb 20, 2008 #10
    yeah thats what I did. K=3.20

    so Kz +Ky*da=3.20z + .2y

    yeah crap forgot to multiply K by da.

    so now that I have the dot, I just integrate?
  12. Feb 20, 2008 #11
    The Kz term shouldn't be there. What's Kz * 0? :p

    You can't just plug in for dA, or just make the integral E * A, because the electric field is non uniform. Also, you should keep everything in terms of variables until the very end.

    Yup, you just integrate. But do you know how to do an area integral?
  13. Feb 20, 2008 #12
    i'm not sure. I know z times y will be the area with z constant and y changing.

    edit is it: [tex]\int_{1} E*dA + \int_{2} E*dA [/tex]
    Last edited: Feb 20, 2008
  14. Feb 20, 2008 #13
    The integrand only depends on y. So integrate over that. But keep in mind that that's only one side of the square you're taking into account.

    For your edit: That looks like how to calculate the net flux through a closed surface. But we're only doing this for one face. What you need to do is set up the integral. What are the limits of integration and what variable are you integrating over?
    Last edited: Feb 20, 2008
  15. Feb 20, 2008 #14
    0 and .250? over y?
  16. Feb 20, 2008 #15
    Yeah but you should really keep .250 in terms of a. Then your final answer will be an expression that you can see the validity of, not just a number.
  17. Feb 20, 2008 #16
    So does this integrand look right or is there no dA involved? sorry for bothering so much.

    [tex] \int^{a}_{0} 0.2y dA [/tex]
  18. Feb 20, 2008 #17
    I don't know where you're getting 0.2 from. The dot product of E and dA was K y dA, which I was trying to lead you into realizing before.

    I don't think I can explain this too well, but: you can't integrate over dA. However, dA = dx*dy. so, what you really need is a double integral:

    [tex]\iint_{square} K y dA = \int_0^a \int_0^a K y dx dy[/tex]

    I dunno if you know how to do this (it's not too hard though; do one integral at a time, starting from the inside one), but that's how I would set up the integral. Since x is constant in the integrand, instead of needing a double integral, you should realize integral is simply this (or this can be figured out by just doing the double integral):

    [tex]a \int_0^a K y dy[/tex]

    Perhaps someone else should explain this more clearly.
    Last edited: Feb 20, 2008
  19. Feb 20, 2008 #18
    .2 was k times A.
    its not your explaining thats the problem, we did no questions like this and there are none in the text.

    I got the right answer now thanks for all your help! know any place I can do some supplimentry reading on this subject? my text book sucks
    Last edited: Feb 20, 2008
  20. Feb 20, 2008 #19
    You can't use A like that, since the electric field is non-uniform.

    A really good book for this stuff is Div, Grad, Curl and All That. It's a lot more advanced that what you need to know for an introductory physics class though but it's going to be really useful later on.

    Last edited by a moderator: Apr 23, 2017
  21. Feb 20, 2008 #20
    yeah I think i'm doing vector calc next year. Im doing Calculas III now. Series and sums and stuff. we do double integrals in a month or so I thin, ill probbaly understand better than.

    thanks for you help
    Last edited by a moderator: Apr 23, 2017
  22. Feb 20, 2008 #21
    I'm thinking that since its a vector integral, the top face of the block should only be affected by 3.20N/C*y [tex]\widehat{k}[/tex]. The other vector runs parallel to the top surface (perpendicular to the surface vector).
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