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Electromotive force

  1. Jan 7, 2009 #1
    1. The problem statement, all variables and given/known data

    A rectangular loop of wire is situated so that one end (height h) is between the plates of a parallel plate capacitor, oriented parallel to the field E. The other end is way outside , where the field is essentially zero. What is the emf in the loop? If the total resistance is R, what current flows? Explain. [Warning: this is a trick question, so be careful if you have invented a perpetual motion , theres probably something wrong with it.
    2. Relevant equations

    V= -[tex]\int[/tex] E[tex]\bullet[/tex] dl

    3. The attempt at a solution


    E=Q/(A*[tex]\epsilon[/tex]0), since each parallel plate capacitor is in the shape of a rectangle, A=hb


    This question comes right out of Griffifth's E&M Textbook and the illustration of the capacitor is on p.294 if you have trouble picturing my description of the capacitor in your head.

    I integrate over h and I get: EMF=Q*ln(h)/(b*[tex]\epsilon[/tex]0)

    Bu the problem says that the field on one end of the parallel plate capicator is 0. Therefore the emf is zero, at least on that end of the capicator is zero.
  2. jcsd
  3. Jan 8, 2009 #2
    maybe if you want some current it is better to use a variable magnetic field in a non-parallel position in respect to the end of the loop wire instead of electric field.
  4. Jan 8, 2009 #3
    wince E=0 zero on the outside of the field , then emf=0. Shouldn't it be that simple?
  5. Jan 8, 2009 #4


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    The Emf is zero, but not for the reason you've stated.

    It is true that [tex]\mathcal{E}=-V=\oint\vec{E}\cdot\vec{dl}[/tex], but is [itex]\vec{E}[/itex] really [itex]\frac{\sigma}{\epsilon_0}\hat{n}[/itex] everywhere between the plates?:wink:
  6. Jan 8, 2009 #5


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    Or, you can look at it this way. This is (apparently) a static problem. So, Kirchhoff (or Gauss) should give you the answer for the emf in (around?) the loop. In other words, there is superfluous information.
  7. Jan 8, 2009 #6
    I know Kirchoff's Laws says the sum of potentials around a loop equal zero. Is Kirchoff's rules the only reason emf =0?
  8. Jan 8, 2009 #7


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    The more fundamental reason is that the field created by a static charge distribution is, not surprisingly, and Electrostatic field. And in chapter 2 of Griffith's (eq 2.19) you should have learned that for any Electrostatic field, the closed path integral:

    [tex]\oint \vec{E}\cdot\vec{dl}=0[/tex]

    because of Coulomb's law dictating a curl-free Electrostatic field.

    Now, what went wrong when you tried to calculate [tex]\oint \vec{E}\cdot\vec{dl}[/tex] in this case?

    Well, a couple of things: First, [itex]h[/itex] is not a variable, so when you integrate over the path of the wire loop, you are not integrating over h. Instead, you might have assumed that outside the capacitor the field was always zero and inside, it was always pointing upward with a magnitude of [itex]\frac{\sigma}{\epsilon_0}[/itex] and so the field is always either zero or perpendicular to [itex]\vec{dl}[/itex] except for the section that is parallel to the field, and so you might have said:

    [tex]\oint \vec{E}\cdot\vec{dl}=\frac{\sigma}{\epsilon_0}\int_0^h dl=\frac{\sigma h}{\epsilon_0}\neq 0[/tex]

    Yikes! Clearly something is still wrong...but what?:wink: ...Well, is the field by the edges of the capacitor really uniform? Aren't there fringing effects to consider?
  9. Jan 8, 2009 #8
    yes. typically , fields outside parallel plate capicators are zero and inside fields are uniform. Wouldn't the fringing field terminate the field inside the capicator once it reaches the edge of the capacitor ?
  10. Jan 8, 2009 #9


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    The field inside a parallel plate capacitor is only exactly uniform if the plates are infinitely large. These plates clearly aren't infinite (although they might be semi-infinite). So the field inside them is only approximately uniform, and only when you are far from the edges. Similarly, the field outside is only approximately zero and only when you are far from the edges. Near the edges, the field is far from uniform, and that is why assuming a uniform field gives you a non-zero answer for the EMF.

    The EMF is zero, and the assumption that the field is uniform everywhere is false.

    Luckily, there is no need to calculate the exact field because all you are interested in is the EMF and eq. 2.19 tells you that must be zero.

    Here's a nice little sketch of the field of a finite capacitor:


    There is a decent explanation of fringing fields here in Section 4.6
    Last edited: Jan 8, 2009
  11. Jan 8, 2009 #10


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    There is no need to make this problem so complicated. You should really try to see the trick, because it is important in E&M. Here's another hint: what do you know about conductors in electrostatic equilibrium. Here's another hint: fringe fields aside, how do the charges tend to arrange themselves in the conducting loop in response to an external electrostatic field.
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