Sheet Resistance Geometry Problem

[tlovoi]

You have a half sphere with a radius of 1mm made up of an insulating material. This insulator is covered by a thin conductive material 1um thick. (1x10^-6 m) This material has a resistivity of 5x10^-8 ohm m. One electrode has a diameter of .5 mm and is centered at the bottom of the half sphere. The other electrode is around the rim of the sphere and makes contact with the conductive film all the way around the "equator" if you will. What is the effective resistance of this material in this geometry?

Half sphere with radius 1x10^-3 meters
Coating Thickness = 1x10^-6 meters
Resistivity = 5x10^-8 ohm m
Diameter of one electrode 5x10^-4 meters

I don't even know where to start. I figure if I knew the resistance of a triangular shape with electrodes on each edge I could break the sphere up into "slices" and integrate treating each slice as a separate resistor in parallel. 1/Rtotal = 1/R1 + 1/R2 etc... but it would require knowing the resistance of a thin film with a triangle shapped geometry.

Can anybody help? Or at least start me in the correct direction...

Thanks!

 

[jbsteele]

The effective resistance of this material in this geometry can be calculated by using Ohm's Law. The resistance of a conductor is equal to its resistivity multiplied by the length of the conductor divided by the cross sectional area. So, in this case, the effective resistance would be: R = 5x10^-8 ohm m * (2π * 1x10^-3 meters) / (.25 * π * (5x10^-4)^2), which works out to 0.1984 ohms.

 

[nlightner]

I can understand your confusion and difficulty in finding a solution to this problem. However, there are some key concepts and equations that can help you approach this problem in a systematic way.

First, we need to calculate the resistance of the thin conductive film. We can use the formula for sheet resistance, which is given by:

Rs = ρ/t

Where Rs is the sheet resistance, ρ is the resistivity and t is the thickness of the film. Plugging in the values given in the problem, we get:

Rs = (5x10^-8 ohm m)/(1x10^-6 m) = 0.05 ohms/square

Next, we need to calculate the resistance of the half sphere. We can use the formula for the resistance of a spherical shell, which is given by:

R = (ρL)/(4πr)

Where R is the resistance, ρ is the resistivity, L is the length of the shell (which is equal to the circumference of the half sphere), and r is the radius of the sphere. Plugging in the values, we get:

R = (5x10^-8 ohm m)(2πr)/(4πr) = 0.05 ohms

Now, we can calculate the effective resistance of the material in this geometry. We can think of the half sphere and the thin conductive film as resistors in parallel. So, the total resistance would be:

Rtotal = (1/Rhalf sphere + 1/Rfilm)^-1

Plugging in the values, we get:

Rtotal = (1/0.05 + 1/0.05)^-1 = 0.025 ohms

Therefore, the effective resistance of this material in this geometry would be 0.025 ohms. I hope this helps you in solving the problem.