How Is Displacement Current Calculated in a Parallel Plate Capacitor?

[PenTrik]

Homework Statement

A sinusoidally-varying voltage V(t) = Vosinωt with amplitude Vo = 10 V and frequency of f = ω/(2π) = 100 Hz is impressed across the plates of a circular-shaped parallel plate air-gap capacitor of radius a = 1.0 cm and plate separation d= 0.01 mm. The amplitude of Maxwell's displacement current ID flowing across the gap between the plates of this capacitor is:

Homework Equations

From what I know, the one equation that we have is
$$I_d=e_0\frac{d\Phi}{dt}$$
Of course, I'm trying to solve for $$I_d$$ here

The Attempt at a Solution

I don't really have much of an attempt here. Sort of lost on how to do it.

 

[ideasrule]

It's good to think back to the reason that displacement currents are needed in the first place. See here: http://en.wikipedia.org/wiki/Displacement_current#To_obtain_the_correct_magnetic_field

Ampere's law states that the integral of B*dl across a closed curve is equal to the current through any surface that you can draw around the curve. As the second diagram in that Wikipedia article's section shows, you can draw two different surfaces, one that the wire passes through and one that no current passes through whatsoever. Displacement current is needed to explain this discrepancy, so displacement current must be exactly equal to the current in the wires (or else the discrepancy would remain).

Another way to calculate displacement current would be to use your equation. You can easily calculate the electric field between between the plates as a function of the total charge on the plate and you know the plates' area, so you can calculate electric flux. Differentiating with respect to time would give you d(phi)/dt.

 

[kerimek]

I would approach this problem by first understanding the physical meaning of Maxwell's displacement current. It is a concept introduced by James Clerk Maxwell in his famous set of equations that describe the behavior of electric and magnetic fields. The displacement current is a term that accounts for the changing electric field in a region of space, and it is important in circuits where there is a time-varying voltage. In this case, the sinusoidally-varying voltage applied to the parallel plate capacitor will create a changing electric field between the plates, resulting in a displacement current.

To solve for the amplitude of the displacement current, we can use the equation you provided, I_d=e_0\frac{d\Phi}{dt}. Here, e_0 is the permittivity of free space, which has a value of 8.85 x 10^-12 F/m. The quantity dΦ/dt represents the rate of change of the electric flux through the capacitor plates. Since the electric field between the plates is uniform, we can calculate the electric flux as Φ = EA, where E is the electric field strength and A is the area of the plates. In this case, the electric field can be calculated using the voltage and plate separation, E=V/d. Therefore, the electric flux can be written as Φ = (V/d)A.

Plugging this into the equation for displacement current, we get I_d = e_0\frac{d\Phi}{dt} = e_0\frac{d}{dt}\left(\frac{VA}{d}\right) = \frac{e_0AV}{d}\frac{d}{dt}\left(sin(\omega t)\right) = \frac{e_0AV\omega}{d}cos(\omega t)

Substituting in the given values, we get I_d = \frac{(8.85\times10^{-12} F/m)(π(0.01m)^2)(10V)(2π(100Hz))}{0.00001m}cos(2π(100Hz)t) ≈ 5.55 x 10^-9 cos(2π(100Hz)t) A.

This is the amplitude of the displacement current flowing across the capacitor plates. It is important to note that this is a time-varying quantity, as it depends on the frequency of the applied voltage. It is also relatively small, which