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I Time-Varying Electric Field in Parallel Plate Capacitor

  1. Sep 16, 2016 #1
    Hi,

    A time-varying (sinusoidal) voltage source is applied to a parallel plate capacitor of length d. Then the E field will vary according to E(t) = V(t)/d. However, this suggests that, for any given time, the E field is constant with respect to spatial coordinates. Therefore, the curl of E is zero.

    The time-varying E field will produce a time-varying B field according to Maxwell's equations. The derivative of B with respect to time will be non-zero. Therefore, the curl of E is non-zero.

    I know that the second statement is correct and the first is incorrect. Why is the first statement incorrect?
     
  2. jcsd
  3. Sep 16, 2016 #2

    mfb

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    For typical capacitor operations, you can neglect dynamic effects - the timescale for those is much shorter than the length of the charging/discharging processes.
    If that assumption is not longer true, you cannot treat the capacitor as single element any more, you have to take its structure into account, and the effects you mentioned become relevant.
     
  4. Sep 16, 2016 #3

    ZapperZ

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    That statement in bold in the quote is VERY odd. It is the same as you saying:

    "For a projectile motion on the way up, since at any given time, the position is a constant with respect to the ground, then it means that dy/dt=0, so the projectile isn't moving and has zero velocity at all times."

    Does that make many sense to you?

    Zz.
     
  5. Sep 16, 2016 #4

    mfb

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    @ZapperZ: The field discussed in post #1 varies in time but not in space. Comparing a field to a projectile doesn't work well.
     
  6. Sep 17, 2016 #5

    vanhees71

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    Well, the apparent contradiction in #1 is that you apply the quasistationary assumption in the first argument, i.e., you neglect the displacement current. The quasistationary equation is then Ampere's Law,
    $$\vec{\nabla} \times \vec{B}=0,$$
    since ##\vec{j}=0##. So there is no contradiction here. If no other sources than the charge on the plates of your capacitor are around, you simply have ##\vec{B}=0##.

    In the next argument, however, you use the full Ampere-Maxwell law including the displacement current, leading indeed to a time-varying magnetic field, but then you also cannot use the quasistatic electric field anymore, because otherwise you indeed contradict Faraday's Law. So then you go beyond the quasistationary (or even quasistatic) approximation, and you have to reconsider the electric field again, which then indeed is not curl free.
     
  7. Sep 17, 2016 #6
    Ok, so if I understand this correctly:

    If a parallel plate capacitor is subjected to a sinusoidal voltage, then the E field between the plates is not actually given by E(t) = V(t)/d, contrary to what my textbook (Fundamentals of Applied Electromagnetics, page 299) states? My textbook uses E(t) = V(t)/d to derive the displacement current, showing that the current in the perfectly conducting wire is equivalent to the displacement current in the perfectly insulating capacitor.

    So, I guess the E field is actually a plane wave in the plane parallel to the capacitor plates, propagating in the direction normal to the plate surfaces? Is that correct?
     
  8. Sep 17, 2016 #7

    mfb

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    That formula is a really good approximation. It is not exact. You see this directly from the missing edge effects as well - the plates don't have infinite sizes.
     
  9. Sep 18, 2016 #8
    Ok, so let's say that the plates have an extremely large area, and the distance between them is small. Then you're saying that E(t) = V(t)/d (where E points from the positive plate to the negative plate) is a very good approximation to the E field near the centre of the plates, far removed from the edges? If that's the case, then the curl of E is essentially 0, so there's no time-varying B field. However, the time-varying E field is associated with a time-varying B field according to the displacement current (εV'(t)/d), so then there is a time-varying B field, leading to a contradiction with the previous statement. I'm still confused here...
     
  10. Sep 18, 2016 #9

    mfb

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    Again: The formula is not exact.

    There are two independent main reasons. One are the edge effects. The other reason is the time-dependence.
     
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