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Easy EM problem

  1. Jan 18, 2007 #1
    1. The problem statement, all variables and given/known data
    For a time independant or static situation

    [tex] -\int_{a}^{b} \vec{E} \bullet d\vec{l} = \Phi(b) - \Phi(a) [/tex]

    a) Show that for any closed path
    [tex] \oint \vec{E} \bullet d\vec{l} = 0 [/tex]

    b)Using Stokes Theorem show that
    [tex] \oint \vec{E} \bullet d\vec{l} = 0 [/tex] implies [tex] \vec{\nabla} \times \vec{E} = 0 [/tex]

    c) [itex] -\int_{a}^{b} \vec{E} \bullet d\vec{l} = \Phi(b) - \Phi(a) [/itex] was derived by integrating [itex] \vec{E} = -\vec{\nabla} \Phi [/itex]. Using Cartesian coordiants show that this implies [itex] \vec{\nabla} \times \vec{E} = 0 [/itex]

    2. The attempt at a solution

    a) For any closed path the endpoints a and b are the same. So
    [tex] -\int_{a}^{a} \vec{E} \bullet d\vec{l} = \Phi(a) - \Phi(a) = 0[/tex]

    Is that it? Just one line?? Is there more i should be considering?

    b) Stokes theorem is

    [tex] \int_{S} \left(\nabla \times v) \bullet da = \oint_{P} v \bullet dl [/tex]

    so here we already know that
    [tex] \oint \vec{E} \bullet d\vec{l} = 0 [/tex]
    thus the right hand side is zero

    [tex] \int_{S} \left(\nabla \times E) \bullet da = 0 [/tex]
    ok since the integral is zero either one of two things - either the integral was over some closed surface S and/or the integrand was zero
    or the curl of E is perpendicular to da

    how would i prove thi formally??

    c) All i do is simply take the curl on both sides of
    [itex] \vec{E} = -\vec{\nabla} \Phi [/itex]
    since the curl of a gradient is zero we prove that it is zero

    since the proof asks to use Cartesian coords i guess i have to do something like this

    [tex] - \nabla \times \left(\frac{\partial \Phi}{\partial x} +\frac{\partial \Phi}{\partial y} +\frac{\partial\Phi}{\partial z} \right) [/tex]
    and prove that the above expression is zero
    Last edited: Jan 18, 2007
  2. jcsd
  3. Jan 18, 2007 #2


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    This is correct
    Well, [tex]0=\oint \vec{E} \cdot d\vec{l}= \int_{S} (\nabla \times \vec{E}) \cdot d\vect{A}\Rightarrow\int_{S} (\nabla \times \vec{E}) \cdot d\vect{A}=0[/tex]

    Since this is for an arbitrary surface, [tex] \nabla \times \vec{E}=0[/tex]

    Yup, try doing that.
  4. Jan 18, 2007 #3


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    Part a is fine. All you need to know about a closed path is that teh start and end points are the same.

    Part b there is only one possibility. The integral over a closed surface of a vector area is zero, but since the vector area is dotted into the curl of the electric field that implies what?

    Part c have you written out [tex] - \nabla \times \left(\frac{\partial \Phi}{\partial x} +\frac{\partial \Phi}{\partial y} +\frac{\partial\Phi}{\partial z} \right) [/tex]
    in full? Try it and see what cancels.

    EDIT: Beaten again. I'm making a habit of this.
  5. Jan 18, 2007 #4

    but why is the curl of E zero as a result??
    i men the integral could be zero if th curl of E is perpendicular to da
    the integral was taken over a closed region
    the curl of E was zero

    how does one imply the other??
  6. Jan 18, 2007 #5


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    The result follows from the fact that there is no restriction on the surface. If curlE, when integrated over any surface is zero, then it must follow that curlE = 0.
  7. Jan 18, 2007 #6
    what do u mean restriction on the surface??
  8. Jan 18, 2007 #7


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    He means the surface can be any surface you like. It is not a defined shape.
  9. Jan 19, 2007 #8
    Note that:

    [tex]\int_{S} (\nabla \times \vec{E}) \cdot d\vect{A} \neq \oint_{S} (\nabla \times \vec{E}) \cdot d\vect{A}[/tex]

    in general. Thus, you could choose a closed surface, but you don't have to. It can be any surface, including one that isn't closed. The only way this can be true for any surface is if the curl of E is 0.

    This is strongly related to the fact that E produces a conservative force.
    Last edited: Jan 19, 2007
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