Why is There No Flux Through the Left Surface of a Discharging Capacitor?

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The discussion centers on calculating the displacement current in a discharging parallel plate capacitor. It highlights confusion regarding the absence of flux through the left surface of the capacitor and the justification for ignoring the charge in the wire. The solution indicates that the displacement current inside the capacitor equals the conduction current, while the external field is negligible due to charge cancellation. Participants emphasize the need to calculate the displacement current as a function of time, potentially using Kirchhoff's laws and differential equations. The conversation ultimately seeks clarity on the relationship between displacement current and conduction current during capacitor discharge.
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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: \oint_S E \cdot dA=\frac{Q_{enc}}{\epsilon_0}
Definition of Displacement Current: I_d=\epsilon_0\frac{d}{dt}\oint_S E \cdot dA

The Attempt at a Solution



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

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 I_d is then just derivative of the rate of change of the charge on the capacitor which is the conduction current.
 
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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).
 
Is my solution correct then?, what about the charge in the wire?
 
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!
 
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?
 
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