Current, Resistivity, and strange questions

[MightyMergz]

Homework Statement

A sheet of aluminum is 1 um thick. A thick copper ring of radius b is placed on the film, and a (smaller) copper disk, of radius a, is placed on the film at the centre of the ring. The disk and ring are connected to the terminals of a battery, so that current I flows through the film from the ring to the disk. Derive, in terms of I, a, b, and the thickness t and resistivity $$\rho$$ of the film, expression for the current density and electric field in the film as functions of the distance r from the centre of the disk and ring. Evaluate the resistance between the disk and ring if b = 10 mm, a = 5mm, t = 1 um.

Homework Equations

current density J is defined as $$J \equiv \frac{I}{A}$$
current density related to the electric field: $$J = \sigma E$$
since $$\sigma = \frac{1}{\rho}$$ therefore $$E = \rho J$$

The Attempt at a Solution

The further out away from the centre, the lower the current density will be, since there will be the same amount of charge flowing over a greater area.

I'm can't think of a way that needs a, b, or thickness. I don't know how the copper is affecting the current, since current is given. All I can think of right now is
$$J(r) = \frac{I}{2 \pi r^{2}}$$
and then $$E(r) = \frac{\rho I}{2 \pi r^{2}}$$
Thanks~

 

[Supernats]

Are you giving the whole problem? What is b?

Edit: My mistake. Sorry.

 

[ogb p]

I would first start by analyzing the given information and breaking down the problem into smaller parts. I would begin by understanding the setup and what is being asked for in the problem. It seems that we are dealing with a sheet of aluminum with a thick copper ring and a smaller copper disk placed on top. The sheet itself has a thickness of 1 um and a resistivity of \rho.

Next, I would use the given equations to derive an expression for the current density and electric field in the film as functions of the distance r from the centre of the disk and ring. Using the definition of current density, J = I/A, we can see that the current density will vary with distance from the centre. As the distance increases, the area over which the charge is flowing also increases, leading to a decrease in current density.

We can also use the equation J = \sigma E, where \sigma is the conductivity and E is the electric field, to relate the current density to the electric field. Since \sigma = 1/\rho, we can rewrite this equation as E = \rho J. This tells us that the electric field will also vary with distance from the centre, as it is directly proportional to the current density.

Now, to incorporate the given parameters of a, b, and t, we can use the fact that the current is flowing from the ring (with radius b) to the disk (with radius a). This means that the current density and electric field will be highest at the ring and decrease as we move towards the centre of the disk. We can express this as J(r) = I/(2 \pi b t) and E(r) = \rho I/(2 \pi b t), where r represents the distance from the centre of the disk.

Finally, to evaluate the resistance between the disk and ring, we can use Ohm's Law (V = IR) and the given parameters to find the voltage difference between the two points. From there, we can use the equation R = \rho l/A to calculate the resistance, where l is the distance between the disk and ring, and A is the area of the aluminum sheet.

In conclusion, the current density and electric field will vary with distance from the centre of the disk and ring, and the resistance between the two points can be calculated using Ohm's Law and the given parameters. The presence of the copper ring and disk will affect the distribution of current and