Time-Varying Electric Field in Parallel Plate Capacitor

In summary: 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?The 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?If a parallel plate capacitor is
  • #1
Seinfeld
4
0
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?
 
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  • #2
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.
 
  • #3
Seinfeld said:
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?

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.
 
  • #4
@ZapperZ: The field discussed in post #1 varies in time but not in space. Comparing a field to a projectile doesn't work well.
 
  • #5
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.
 
  • #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?
 
  • #7
Seinfeld said:
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?
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.
 
  • #8
mfb said:
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.

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...
 
  • #9
Again: The formula is not exact.

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

Related to Time-Varying Electric Field in Parallel Plate Capacitor

1. What is a time-varying electric field in a parallel plate capacitor?

A time-varying electric field in a parallel plate capacitor is an electric field that changes in strength or direction over time. This can occur when the voltage across the capacitor is changing, causing the electric field between the plates to also change.

2. How does a time-varying electric field affect the capacitance of a parallel plate capacitor?

A time-varying electric field can change the capacitance of a parallel plate capacitor. This is because the capacitance is directly proportional to the electric field strength, so as the field changes, the capacitance changes as well.

3. What is the relationship between the time-varying electric field and the current in a parallel plate capacitor?

The time-varying electric field can induce a time-varying current in a parallel plate capacitor. This is known as displacement current and is related to the rate of change of the electric field with time.

4. How does the geometry of a parallel plate capacitor affect the time-varying electric field?

The geometry of a parallel plate capacitor, specifically the distance between the plates and the surface area of the plates, can affect the strength of the time-varying electric field. A smaller distance between the plates or a larger surface area will result in a stronger electric field.

5. What are some real-world applications of time-varying electric fields in parallel plate capacitors?

Time-varying electric fields in parallel plate capacitors are commonly used in electronics, such as in radio frequency circuits, where they can be used for filtering, tuning, and coupling signals. They are also used in power systems for voltage regulation and in medical devices for imaging and therapy.

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