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The electric line integral

  1. Feb 17, 2010 #1
    In some books I have seen:

    [tex]\oint \mathbf{E} \cdot d\mathbf{s}=0[/tex]

    Since the Electric Field is meant to be conservative.

    Elsewhere, however, I have also seen:

    [tex]\oint \mathbf{E} \cdot d\mathbf{s} = -\frac{d\Phi_B}{dt}[/tex]

    What's going on here?

  2. jcsd
  3. Feb 17, 2010 #2


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    The first case is for electro-statics (no changing fields) and hence is true for the special case when [tex]\frac{d\Phi_B}{dt}=0[/tex]
  4. Feb 17, 2010 #3
    Ah, thanks Matterwave
  5. Feb 17, 2010 #4
    This is true, of course.

    What is amazing is that circuit or physics text books often quote the special case erroneously. They sometimes even call it a version of Kirchoff's Voltage Law, which is not correct. Any AC circuit that includes an inductor violates the quoted special case. Students should keep an eye out for this, and make sure the special case is really valid.
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