# Maxwell's Equations and a circular capacitor

1. Oct 22, 2011

### TheTourist

1. The problem statement, all variables and given/known data

The current I=I0exp(-t) is flowing into a capacitor with circular parallel plates of radius a. The electric field is uniform in space and parallel to the plates.

i) Calculate the displacement current ID through a circular loop with radius r>a from the axis of the system

ii) Calculate an expression for the electric field between the capacitor plates.

3. The attempt at a solution

i) As the plates are being charged up by the current I, the displacement current ID is increasing due to a growing electric field between the plates. I therefore reasoned that as the current I is decreasing with time, the displacement current is increasing with time by an equal amount so the answer would be ID=I0exp(t).
Though I don't know if this is correct.

ii) I used Gauss's law in integral form and a cylinder around the positive capacitor plate.
$\int$E.dS=Q(t)/ε0

I performed the integration using the top surface of a cylinder, RdRd$\phi$$\hat{z}$

and ended with the equation E(t)=Q(t)/(ε0πa2)$\hat{z}$
where a=R and I=dQ/dt
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Oct 22, 2011

### Matterwave

It is true that the Electric field is increasing, but why would that imply an increase in the displacement current? Remember that the displacement current is proportional to the time derivative of the Electric field, so that "the electric field increasing" only implies that the displacement current is positive. Whether the displacement current is increasing or decreasing depends on the SECOND derivative of the electric field.

Use Ampere's law with a circle as the boundary and try two different surfaces bound by that circle, one which intersects with the wire and one which goes through the gap in the capacitor. The answer (the strength of the magnetic field) has to be the same from the two surfaces. What does that tell you about the displacement current?

3. Oct 25, 2011

### TheTourist

I think I have been misunderstanding the physical meaning of the displacement current. You have helped me to think about it in the correct way, thank you Matterwave.
The displacement current should be equal to the current in the wire in this case, i.e. amperes law gives the same result inside the capacitor as the wire, even though there is no charge flow in the capacitor.

4. Oct 25, 2011

Yep. =D