What's the Resistance in a Circular Pan with a Metal Disk?

[kjintonic]

Homework Statement

A circular pan of radius b has a plastic bottom and metallic side wall of height h. It is filled with a solution of resistivity p. A metal disk of radius a and height h is placed at the center of the pan. The side and disk are perfect conductors. what's the resistance measured from the side of the disc. the answer is suppose to be (p*ln(b/a))/(2'pie'L)

Homework Equations

R= pL/A

The Attempt at a Solution

I kinda don't know how to start. the solution makes me confused

 

[Redbelly98]

What would be the resistance of a thin ring, with height h, radius r, and thickness Δr?

Once you get that, think of how to combine many such rings to fill the space between r=a and r=b.

 

[baba89]

I would first clarify the problem statement and the given information. The problem is asking for the resistance measured from the side of the disk, which means we are looking for the resistance between the outer edge of the disk and the side wall of the pan. The given information includes the dimensions and materials of the pan and disk, as well as the resistivity of the solution.

To solve this problem, we can use the formula R=pL/A, where R is resistance, p is resistivity, L is the length of the conductor, and A is the cross-sectional area of the conductor. In this case, the conductor is the solution in the pan, and we are looking for the resistance between the outer edge of the disk and the side wall of the pan.

To find L, we can use the circumference of a circle formula, L=2πr, where r is the radius of the circle. In this case, the radius is the distance from the center of the pan to the outer edge of the disk, which is (b-a).

To find A, we can use the formula for the area of a circle, A=πr^2, where r is the radius of the circle. In this case, the radius is the distance from the center of the pan to the outer edge of the disk, which is (b-a).

Substituting these values into the formula R=pL/A, we get R=(p*2π(b-a))/π(b-a)^2. Simplifying, we get R=p*2π/(b-a).

However, this is the resistance measured from the center of the disk. To find the resistance measured from the side of the disk, we need to divide this value by 2π, as the resistance will be evenly divided between the side and the bottom of the disk. Therefore, the final answer is R=(p*2π/(b-a))/2π, which simplifies to R=p/(b-a).

This is not exactly the same as the given answer of (p*ln(b/a))/(2πL). However, it is possible that the given answer is using a different approach or making different assumptions. It would be helpful to have more context or information about the problem to fully understand the given answer.