Finding displacement current in an AC capacitor circuit

[Peter Andrews]

Homework Statement

A capacitor is made of two parallel plates of area A, separation d. It is being charged by an AC source. Show that the displacement current inside the capacitor is the same as the conduction current.

Homework Equations

I_disp = ε(dΦ_E/dt)
Q = CV
C = Aε/d
X_c = 1/(2πƒC)
Q(t) CV(1-e^-t/τ)

The Attempt at a Solution

First of all, as it's an AC circuit so we won't be completely charging the capacitor, so I don't think we will use the exponential charging equation for a capacitor.
The displacement current can be obtained by differentiating Q=CV.
That gives (dQ/dt) = C(dV/dt). Assuming V = V_osin(2πft), we get a suitable expression for I_d. This leads to a current answer. But of we want to use the electric flux equation then how can we do that?

 

[cnh1995]

But of we want to use the electric flux equation then how can we do that?

What is the equation for electric flux in terms of charge Q?

 

[Peter Andrews]

What is the equation for electric flux in terms of charge Q?

Can use the Guass theorem, q_enclosed/ε = ∫E.ds (closed integral, I don't know how to insert that sign). But what Guassian surface to choose?

 

[cnh1995]

Can use the Guass theorem, q_enclosed/ε = ∫E.ds (closed integral, I don't know how to insert that sign). But what Guassian surface to choose?

You can assume the electric field to be uniform. So, electric flux will be simply electric field*area. What is the relevant formula for electric field here?

 

[Peter Andrews]

You can assume the electric field to be uniform. So, electric flux will be simply electric field*area. What is the relevant formula for electric field here?

But how can we assume electric field to be constant when the charge is changing? The electric field in a capacitor is σ/ε. σ arial charge density. If we use that, then:
Φ = A.E
= Q/ε
Now differentiating,
(dΦ/dt) = (DQ/dt)/ε
So ε(dΦ/dt) = (DQ/dt)
Hence I_displacement is dq/dt following which I come back to the process given in my solution above. Is there no other way via which I can get the result, without having to apply dq/dt on capacitor?

 

[cnh1995]

Φ = A.E
= Q/ε

Right.
The field is changing but it is uniform at any instant.

 

[Peter Andrews]

Right.
The field is changing but it is uniform at any instant.

Exactly. Constant and uniform have different meanings, at least in fields.

 

[cnh1995]

Exactly. Constant and uniform have different meanings, at least in fields.

That is why I immidiately changed it to 'uniform'.