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Displacement Current understanding

  1. Aug 1, 2013 #1
    1. The problem statement, all variables and given/known data
    I am having trouble grasping why exactly displacement current and conduction current are sometimes equal to each other and sometimes are not.

    My textbook states that the displacement current is defined as ε0 * d(EA)/dt. That is, epsilon not times change in electric flux. I understand the derivation of this formula, but in some problems I attempt, I have to make use of the fact that displacement current = conduction current (such as, for instance, when I am told to find the displacement current density in a wire and I am only given conduction current). In other problems I have attempted, I've found different (yet correct) values of displacement current and conduction current.

    Thanks in advance.

    2. Relevant equations

    Ic = Id = ε0 * d(EA)/dt.
     
  2. jcsd
  3. Aug 2, 2013 #2

    vanhees71

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    What textbook is this?

    Anyway, the socalled discplacement current is a term in the Ampere-Maxwell Law, which is one of the fundamental Maxwell equations of electromagnetism, which reads in the vacuum:
    [tex]\vec{\nabla} \times \vec{B}=\mu_0 \epsilon_0 \partial_t \vec{E}+ \mu_0 \vec{j}.[/tex]
    Here [itex]\vec{E}[/itex] and [itex]\vec{B}[/itex] are the components of the electromagnetic field and [itex]\vec{j}[/itex] is the current density. Only the latter is caused by moving charges, and the conduction current in a conductor is one example for such a current. The "discplacement current" is just the time derivative of the electric field. It was added by Maxwell to the older Ampere Law for consistency reasons.

    The modern way to understand electrodynamics is quite different. There one starts from the invariance of natural laws under Poincare transformations, i.e., the relativistic space-time structure and derives field equations from Hamilton's principle of least action. It turns out that a massless spin-1 field properly describes the electromagnetic field leading to Maxwell's equations. In this approach, the "displacement current" comes simply out from the Poincare invariance of the field equations. It should be put to the left-hand side of the equation and not be interpreted as a "current" in the sense of moving charge.
     
  4. Aug 2, 2013 #3
    Young Freedman University Physics

    The textbook uses a wire connected to a parallel plate capacitor in order to derive the formula for displacement current.

    C = capacitance, Q = charge, V = voltage applied

    CV = Q. Voltage = Ed, where d = distance between plates, since E (the electric field) is uniform.

    CEd = Q.The capacitance of a parallel plate capacitor is (ε0*A)/d, where ε0 is the permittivity of free space. So substituting in this for C:

    E*A*ε0 = Q. Now, since E*A is electric flux, EA = Q/ε0.

    Now, the book takes a time derivative of both sides, and does the following:

    d(EA)/dt = 1/ε0 * dQ/dt. The book states that dQ/dt is the conduction current (Ic).

    Ic = ε0 * d(EA)/dt. Now the book pulls this idea of displacement current out of thin air and sets it equal to this exact quantity. So for many problems I assumed that they are both equal by definition, but in some of them this assumption has led to incorrect answers. So I am wondering how displacement current is defined to be equal to conduction current yet sometimes it isn't.
     
  5. Aug 6, 2013 #4

    rude man

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    Conduction current is never the same thing as displacement current. In a region of displacement current there are no charges moving in the region defined as the displacement current region. So for example the current between the plates of a capacitor is a displacement current whereas the current in the capacitor wires is a conduction current. And yes, the two currents are the same since it's the same current loop and charge cannot accumulate in any node along the loop.
     
  6. Aug 6, 2013 #5

    Andrew Mason

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    The displacement current is not a current at all, of course. It is added to Ampere's law in order to avoid a discontinuity of the magnetic field between the plates of a capacitor. Prof. Lewin has a good explanation of it here on Youtube.

    AM
     
  7. Aug 6, 2013 #6

    rude man

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    Also, displacement current generates a magnetic field (see Ampere's law) exactly the same as a conduction current.
     
  8. Aug 6, 2013 #7

    vanhees71

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    One should interpret [itex]\vec{E}[/itex] and [itex]\vec{B}[/itex] as the 6 components of the one and only electromagnetic field. In the inhomogeneous Maxwell equations one should write everything connected to the em. field to the left and everything concerning the sources to the right:
    [tex]\vec{\nabla} \cdot \vec{E}=\rho, \quad \vec{\nabla} \times \vec{B}-\frac{1}{c} \partial_t \vec{E}=\frac{1}{c} \vec{j},[/tex]
    where I used the physically more sensible Heaviside-Lorentz (i.e., rationalized Gauß units).

    In the relativistic covariant form this is one equation for the Faraday tensor, [itex]F_{\mu \nu}[/itex], i.e.,
    [tex]\partial_{\mu} F^{\mu \nu}=\frac{1}{c} J^{\nu}.[/tex]
    This makes the physical meaning of the various terms in the Maxwell equations clear!

    The interpretation of the [itex]\partial_t \vec{E}[/itex] term in the Maxwell-Ampere Law as another source for the magnetic field is due to the then not yet known relativistic nature of electromagnetism and shouldn't be taught anymore, over 100 years after Einstein and Minkowski!
     
  9. Aug 6, 2013 #8
    Reminder that the OP listed this in "Introductory Physics Homework" :)
     
  10. Aug 6, 2013 #9

    vela

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    Can you provide us with a specific example of when you found they weren't equal?
     
  11. Aug 6, 2013 #10

    rude man

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    Motion seconded!
     
  12. Aug 7, 2013 #11

    vanhees71

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    I disagree. The correct concepts are the more important in the introductory course. One should not learn old fashioned concepts in highschool or undergrad studies which you then have to correct later! It's much more difficult to forget wrong concepts than to learn the right ones from the very beginning!
     
  13. Aug 7, 2013 #12

    D H

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    No, it's not. A freshman reading University Physics does not have the mathematical background to understand what you posted earlier. This why science curricula are the way they are.

    A couple of useful quotes:

    Ian Stewart and Jack Cohen:
    A lie-to-children is a statement that is false, but which nevertheless leads the child's mind towards a more accurate explanation, one that the child will only be able to appreciate if it has been primed with the lie.​

    Isaac Asimiov:
    When people thought the earth was flat, they were wrong. When people thought the earth was spherical, they were wrong. But if you think that thinking the earth is spherical is just as wrong as thinking the earth is flat, then your view is wronger than both of them put together.​


    So let's get back on topic, but please keep this thread to the level of an introductory freshman/sophomore physics class.
     
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