Calculate the the displacement

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In summary, the conversation discusses the problem of calculating the displacement current in a parallel plate capacitor as it discharges. The Gauss Law and the definition of displacement current are used to construct a Gaussian surface and determine the flux through it. The conversation also addresses the confusion regarding the flux through surfaces outside the capacitor and the role of charges in the wire. The conclusion is that the displacement current outside the capacitor is zero and inside the capacitor it is equivalent to the electric current in the wires. The conversation also mentions the need to calculate Id=Ic and solve a differential equation to determine the displacement current at a specific time.
  • #1
asymptotically
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Homework Statement



Hi, the problem is to calculate the the displacement current flowing through a surface that lies in between the two plates of a parallel plate capacitor as the capacitor discharges. I think I have a correct solution but I'm not 100% sure on why it's correct.


Homework Equations



Gauss Law: [itex]\oint_S E \cdot dA=\frac{Q_{enc}}{\epsilon_0} [/itex]
Definition of Displacement Current: [itex]I_d=\epsilon_0\frac{d}{dt}\oint_S E \cdot dA[/itex]

The Attempt at a Solution



Constructing a Gaussian surface about one of the plates like this we have [itex]\oint_S E \cdot dA=\frac{Q(t)_C}}{\epsilon_0} [/itex].

Now this is the part I'm slightly confused about, the flux through every surface except the surface
that lies in between the plates has to be zero, why is there no flux through the left surface. Also we also only consider the charge on the capacitor and ignore the charge in the wire? Are these simplifications or is there justification for this.

The answer for [itex]I_d[/itex] is then just derivative of the rate of change of the charge on the capacitor which is the conduction current.
 
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  • #2
The total current is conserved, that's the meaning of the displacement current.
The displacement current outside the capacitor vanishes because the field vanishes.
The displacement current inside the capacitor is therefore the electric current in the wires.

The field outside the capacitor is the superposition of the field produced by opposite charges.
Therefore is is zero (or almost zero).
 
  • #3
Is my solution correct then?, what about the charge in the wire?
 
  • #4
Yes the "Relevant equation" is correct, but it only helps you to prove you last sentence.
The wire has no charges, the charges accumulate on the plates (can you explain why?).
You still need to calculate Id=Ic!
 
  • #5
maajdl said:
Yes the "Relevant equation" is correct, but it only helps you to prove you last sentence.
The wire has no charges, the charges accumulate on the plates (can you explain why?).

Nope, why do wires need to be covered in an insulator if they don't have a charge?

maajdl said:
You still need to calculate Id=Ic!

Na the problem just say's Ic=I at some time, what is the displacement current at this time, to calculate Ic as a function of t I assume I just use one of kirchhoffs law and solve a Diff Eq?
 

1. What is displacement?

Displacement is the distance and direction of an object's change in position from its starting point.

2. How do you calculate displacement?

Displacement can be calculated by subtracting the initial position from the final position of an object

3. What are the units of displacement?

The units of displacement are typically represented in meters (m) or any other unit of length.

4. Can displacement be negative?

Yes, displacement can be negative if the final position is behind the initial position. This indicates that the object has moved in the opposite direction of its initial position.

5. What is the difference between displacement and distance?

Displacement takes into account the direction of an object's movement, while distance only calculates the total amount of ground covered by the object.

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