# Magnetism, Displacement Currents and Parallel Plate Capacitors

## Homework Statement

The preceding problem was an artificial model for the charging capacitor, designed to avoid complications associated with the current spreading out over the surface of the plates. For a more realistic model, imagine thin wires that connect to the centres of the plates (see figure (a)). Again, the current I is constant, the radius of the capacitor is a, and the separation of the plates is w << a. Assume that the current flows out over the plates in a such a way that the surface charge is uniform, at any given time, and is zero at time t = 0.

(a)

Find the electric field between the plates, as a function of t.

I=c dV/dt∴Idt=Cdv

(b)

Find the displacement current through a circle of radius s in the plane midway between the plates. Using this circle as your "Amperian loop", and the flat surface that spans it, find the magnetic field at a distance s from the axis.

(c)

Repair part b, but this time uses the cylindrical surface in figure (b), which is open at the right and extends to the left through the plate and terminates outside the capacitor. Notice that the displacement current through this surface is zero, and there are two contributions to the enclosed current.

Not sure

## The Attempt at a Solution

I'm trying to do part (a), but I am not sure of an equation which has time in it. So far, I have
got:

$$I = c\frac{dv}{dt}$$

Thus

$$Idt = Cdv$$

But I am not sure if this is right.

Is it? and if so, what might be some useful equations/rules to use?

Thanks in advanced,

TFM

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## Answers and Replies

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I have done some more for part a):

$$\oint_S E \cdot da = \frac{Q}{\epsilon_0}$$

For a capacitor, Q = CV

from the equatuion I posted in the first post,

$$C = I \frac{dt}{dv}$$

putting together:

$$\oint_S E \cdot da = \frac{vI\frac{dt}{dv}}{\epsilon_0}$$

Gives:

$$EA = \frac{vI\frac{dt}{dv}}{\epsilon_0}$$

$$A = \pi w^2$$

giving E:

$$E = \frac{vI\frac{dt}{dv}}{\epsilon_0 \pi w^2 }$$

Does this look right?

TFM