Resistivity calculation from two-probe resistance measurement of sheet

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Calculating resistivity from two-probe resistance measurements requires consideration of contact area geometry, particularly when using point contacts, which theoretically yield infinite resistance. The discussion highlights that the perimeter of each contact area is crucial for accurate calculations. An exact solution can be derived using elliptical equipotentials around each contact area, with the central ellipse acting as the bisector between them. The potential distribution is related to the logarithmic ratio of distances to the centers of these ellipses. Understanding these factors is essential for accurate resistivity calculations.
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Homework Statement
You make a two-probe resistance measurement (probe spacing S) of an infinite sheet (thickness t) of a high-resistance material. How do you calculate the resistivity of the material?
Relevant Equations
R = ohm, rho = ohm*m
I'm really not sure. Obviously I can get the units right with Resistance * thickness, but I assume there's a correction factor here that I can't find anywhere?
 
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I studied this once out of my own interest. I concluded that as stated in post #1 it makes no sense. What you also need to know is the perimeter of each contact area. With point contacts the theoretical resistance is infinite.
If we take those areas to be elliptical, we can have an exact solution with nested elliptical equipotentials around each contact ellipse. The 'central ellipse', i.e. where the one family transmutes into the other, is the perpendicular bisector of the two contact areas. The potential at any point is proportional to the logarithm of the ratio of distances to the 'centres', i.e. where the two families of ellipses would shrink to zero.
 
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Thread 'Chain falling out of a horizontal tube onto a table'
My attempt: Initial total M.E = PE of hanging part + PE of part of chain in the tube. I've considered the table as to be at zero of PE. PE of hanging part = ##\frac{1}{2} \frac{m}{l}gh^{2}##. PE of part in the tube = ##\frac{m}{l}(l - h)gh##. Final ME = ##\frac{1}{2}\frac{m}{l}gh^{2}## + ##\frac{1}{2}\frac{m}{l}hv^{2}##. Since Initial ME = Final ME. Therefore, ##\frac{1}{2}\frac{m}{l}hv^{2}## = ##\frac{m}{l}(l-h)gh##. Solving this gives: ## v = \sqrt{2g(l-h)}##. But the answer in the book...

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