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Displacement Current Concept

  1. May 14, 2006 #1
    I have a question in regards to displacement current. Is this theory mainly used to describe the field when a capacitor has a shape other than a flat one?

    Can anyone link me a site that clearly explains how we can use the concept of displacement current?

    From what I understand through displacement current is how we can see that charge flows from one plate to the other in a circuit?(or theory of since displacement current is not real?)

    I'm confused :)
  2. jcsd
  3. May 14, 2006 #2
    I have a really dopy (but insightful yet simple) way of looking at displacement current.

    We know that currents produce B fields with a curl in the direction of the current, correct. This means that the B field circulates the current, as indicated by the right hand rule.

    Now, say we have a constant electric field in some dielectric. since the E-field is constant all of the atoms in the dieletric will be polarized in the direction of the E field (so if the material is already polar then a torque will act on its dipole aligning it with the E-field or is it is not polar then it will become polar and the new dipole is aligned with the e-field). In this constant e-field the polarized atoms in the dielectric are charged but they are not moving, correct? Now, let us change the E-field (like the e-field in a capacitor in an AC circuit). What happens to the polarized atoms in the dielectric under this condition? The bound charges move around! This movement of charge in turn induces a b-field, which is why that displacement current density term (derivative of D wrt t) is tacked onto Maxwell's 4th equation!

    So, yes, this is in a sense how charge moves through the dielectric of a capacitor and why the displacement current is only caused by changing D fields.

    Displacement current is a VERY real thing. It is the current caused by the movement of dipoles in a dielectric, and this movement of dipoles is caused by a CHANGING D field and notice how Maxwell's 4th equation nicely accounts for this.
    Last edited: May 14, 2006
  4. May 14, 2006 #3
    Thanks!!! That helped alot!
  5. May 14, 2006 #4


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    Staff: Mentor

    Here's how I justify the displacement current physically:

    Start with a very long straight current-carrying wire. We can calculate the magnetic field around the wire by using Ampere's Law (without the displacement current term) in the usual way, by assuming cylindrical symmetry and setting up a circular Amperian loop around the wire.

    Now put a short break in the wire, and attach the cut ends to the centers of two circular metal plates oriented perpendicular to the wire. We've just constructed a circular parallel-plate capacitor. We can put a dielectric inside if we want, but it's not necessary for this argument.

    Suppose we have a current in the two sections of the wire, just like before. Charge must be building up on the plates, so the current can't continue forever, but a short period of time is all we need. While the current is flowing, we can calculate the magnetic field around the two sections of wire just like we did before, using Ampere's law (without the displacement current term).

    But what about the magnetic field around the space between the plates? There's no current in that space, so Ampere's law (without the displacement current term) gives zero for the magnetic field. But surely the magnetic field doesn't drop off sharply to zero, and then rise suddenly to its original value, as we move past the capacitor. There has to be continuity here. If the plates are close together, the magnetic field shouldn't change very much as we go past the capacitor.

    This leads us to invent a fictitious current between the plates, which has a value equal to the "real" current in the wires. Now suppose the real current is constant. Then the charge on the plates increases linearly, and so does the electric field between the plates. Therefore the required fictitious current (which is of course the displacement current) must be proportional to the rate of change of the flux of E between the plates. Using this circular parallel-plate example, we can find out what the proportionality constant is.

    You don't need a dielectric in order to have a displacement current. A vacuum-filled capacitor will do just fine.
  6. May 14, 2006 #5
    this is true.
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