Resistance of infinite sheet

In message , Joel writes

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Try to visualize a lattice of unit resistance's. This gives you lots of resistance's in a series/parallel arrangement. You should see that for a very large sheet you will get a large resistance. Take the limit...

You have it basically right. The resistance perpendicular to a radius of your 'ring' would be the resistivity, times the radial thickness of the ring, divided by the area presented in the radial direction. If the thickness of your sheet of material is 'T', then the resistance of a ring with ID of r1 and OD of R2 would be.... R = (rho / (2pi T)) * int[r1-r2] (dr/r) R = (rho / (2pi T)) * [ln(r2) - ln(r1)]

Letting r2 approach infinity and leaving r1 unspecified, you'd be tempted with R = (rho / (2pi T)) * [ln(infinity) - ln(r1)] R = (rho / (2pi T)) * (infinity - ) = infinity

But, as you may have noticed, with r1 set to zero, you get R = (rho / (2pi T)) * (ln(infinity) - ln(r1)] with r1 -> 0 as a limit Simple evaluation of the limit would result in... R = (rho / (2pi T)) * [infinity - (-infinity)] R = (rho / (pi T)) * [infinity] R = infinity

daestrom

Set this problem up with finite L and W. Integrate, then look at the limiting resistance as L or W (or both)go to infinity.

Hint: your problem specification seems inconsistent. If L and W are both infinite, the resistance will be infinite.