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Curl and Divergence

  1. Sep 18, 2004 #1
    Hi All,

    Given electric field E=c(2bxy,x^2+ay^2), I need to determine the constants a and b such that CURL E = 0 and DIV E = 0. I'm also given a path from (0,0) , (1,0) and (1,1).

    Ok so the curl = 0+0+cx(2-b) = 0
    and the divergence = 2cy(b+a) = 0

    How do I solve for a and b at this stage given so many variables?

    Also how do I compute the potential in this case.
  2. jcsd
  3. Sep 18, 2004 #2

    Doc Al

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    First, recheck your expression for the curl.

    Then ask yourself: what must "b" be to make the curl always zero (that is, for any value of x and y)? What must be the relation between "b" and "a" to make the divergence always zero?
  4. Sep 18, 2004 #3
    ok so I redid the curl and got 2xc(1-b) = 0

    therefore b=1 to satisfy that the curl = 0
    and a=-1 so that the div = 0 :)

    ok that was easy after all.......

    now how do I compute the potential?
  5. Sep 18, 2004 #4


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    Given potential [tex]\Phi[/tex] you must have:
    How can you use these equations to determine [tex]\Phi[/tex] ?
  6. Sep 18, 2004 #5
    Remember also that when you integrate with respect to one variable, the "constant" will actually be a function of the other variable.

    Actually, I tried working this problem out for you. But the answer I kept getting was slightly off from what the real potential should be (sorry, it's been a couple years since I took multivariable). In any case, the answer should be f(x,y) = (1/3)acy³ + cx²y. But definitely do arildno's method and make sure that you get this answer (professors don't react to well to answers without work, as I'm sure we've all experienced).
    Last edited: Sep 18, 2004
  7. Sep 18, 2004 #6


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    Notice that by differentiating the first equation with respect to y, [itex]\frac{\partial^2\Phi}{\partial x\partial y}= 2bcx[/itex] and by differentiating the second equation with respect to x, [itex]\frac{\partial^2\Phi}{\partial y\partial x}= 2cx[/itex]. Since, for any function with continuous second partials, the two mixed partial derivatives must be equal this is possible only if b= 1. Otherwise there is no [itex]\Phi(x,y)[/itex] that has those derivatives.

    Integrating the first equation with respect to x, treating y as a constant, [itex]\Phi= bcx^2y+ f(y)[/itex] (the "constant" of integration may depend on y).

    Differentiating THAT equation with respect to y, we get [itex]\frac{\partial\Phi}{\partial y}= bcx^2+ \frac{df}{dy}[/itex].

    Comparing that with the second given equation, we must have
    [itex] bcx^2+ \frac{df}{dy}= cx^{2}+acy^{2}[/itex].

    In order for that to be true, the terms involving x must cancel which happens only for b= 1. Taking b= 1 we can cancel that term and have
    [itex]\frac{df}{dy}= acy^2[/itex]. Integrating that, [itex]f(y)= \frac{1}{3}acy^3+ C[/itex]. Because f depends only on y, C really is a constant.
    Putting those together, we have:
    [itex]\Phi(x,y)= cx^2y+\frac{1}{3}acy^3+ C[/itex]
    as long as b= 1. If b is not 1, there is no such function.
    Last edited: Sep 18, 2004
  8. Sep 18, 2004 #7


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    Since the first part of the exercise was to determine those values of a and b which gives zero curl(&divergence,) those are the ones that were implied to be used in my example.
    (The existence of the potential has therefore been assured by these choices)
  9. Sep 18, 2004 #8
    That's why I kept messing up when I tried to actually calculate the potential! I forgot that b must equal 1.

    Also I completely forgot about Clairaut's theorem. Amazing how much stuff there is to forget in math.
  10. Sep 18, 2004 #9
    Thanks guys....

    Arunma, how did you know the solution without being able to derive it correctly?
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