how to compute( without using dmm) a hollow cuboid's resistance between two points on same face but not lying on the edges. the line joining 2 points is parallel to the length of the cuboid.
resistivity is uniform
another related doubt - will resistance between those same points change if the cuboid is deformed (the box is crumbled)
edit:contact points arent "dimensionless ", i mean copper wires are connected to cuboid at A and B
Resistance depends on the resistivity of the material and the combination of the cross-section of the material and the distance between the test points.
You can easily imagine that a thick wire will have lower resistance than a thin wire of the same length. And a shorter wire will have less resistance than a longer wire of the same cross-section.
My maths is not good enough to do the necessary calculations for your cube. Use a multimeter 
I think it won't make any difference if the cube is deformed as long as the deformation does not result in parts touching which did not previously touch.
...R
sorry click on wrong button. How do I delete this?
The shape of the insulator or conductor has no bearing on its qualities. Since voltage travels on the outside and not the inside.
As Robin2 said a multi meter will tell you all you need to know. You can change the shape all you want so long as to length of the distance that your are measuring doesn't change it will read the same.
As seen in the picture the insulator has lots of cup like shapes, Why because the voltage travels on the outside. So you take an insulator that is three feet long , squish it down so that it is one foot long and the outside is not touching any surface of it self. the distance from A to B is the same electrically speaking.
SumitAich:
How to compute (without using dmm) a hollow cuboid's resistance between two points on same face but not lying on the edges. The line joining 2 points is parallel to the length of the cuboid. Resistivity is uniform.
Contact points aren't "dimensionless."
Take a look at how hard it is to find the resistance of a solid sphere between electrodes at the poles (which, on the face of it, seems like it ought to be a relatively simple problem): (PDF) The Notion of Electrical Resistance | Carlos E. Solivérez - Academia.edu. Judging from that, I'd bet that an analytic solution to your problem will be......exceedingly difficult.
A shell should be easier than a solid object, I think...
Everything I look at (or remember) implies that you do some pretty heavy-duty calculus based on the Electric Field equations to calculate the current densities, and then work backward via Ohms law to get an equivalent resistance.
Where are you getting these "interesting" problems?
ok
You could simplify the problem by imagining that each side of your cube is a matrix of resistors (whose resistance is equal to the resistance of the equivalent part of the "real" cube). Then use Ohm's law and Kirchoff's law to solve for the equivalent resistance between your chosen points (google solving complex resistor networks).
DaveEvans:
You could simplify the problem by imagining that each side of your cube is a matrix of resistors (whose resistance is equal to the resistance of the equivalent part of the "real" cube). Then use Ohm's law and Kirchoff's law to solve for the equivalent resistance between your chosen points (google solving complex resistor networks).
holy mother of God that f**kin worked! thanks a lot
SumitAich:
that f**kin worked! thanks a lot
Sorry, but I can't resist asking How do you know?
...R
Less than three hours between my post and "it works!" Hmmm. If the OP solved the problem in that amount of time, he or she is quite the mathematician (at least compared to yours truly, which, granted, isn't saying much). However, my guess is that it's sarcasm, or a "yeah, right." But, without "emoticons," it's hard to say. I am curious...
Hi,
I googled a hollow cuboid resistance between two points on same face but not lying on the edges. the line joining 2 points is parallel to the length of the cuboid.
You would be surprised how many times it has been asked with same diagram.
Tom.... :o
The resistance of a square of material, measured between its edges, is constant. The bigger you make the square, the longer the path and the wider the path, so the two cancel out. The same is true if you measure the resistance between opposite corners.
I suspect that, with a solid cube, the resistance decreases the larger the cube. There is more material for the current to flow through.
I also suspect that a hollow cube would act similarly to a square, so, in that case, size doesn't matter.
The further apart your contacts, the greater the direct path between them, but lesser distance in any other paths.
I don't know the answer, but hope that helps.
DaveEvans:
Less than three hours between my post and "it works!" Hmmm. If the OP solved the problem in that amount of time, he or she is quite the mathematician (at least compared to yours truly, which, granted, isn't saying much). However, my guess is that it's sarcasm, or a "yeah, right." But, without "emoticons," it's hard to say. I am curious...
feels good to be back..
using your analogy of resistor network, i came up with this approach..
consider current injected into A through B , we'll have equipotential (potential symmetry) points on opposite sides of line AB .
joining them together we have a square lamina. rest is piece of cake.
I can picture the symmetry in two special cases, but not generally. I see symmetry if:
A and B are on a line that bisects a face and is parallel to an edge, where A and B are equidistant from the middle of the face,
or if A and B are on an edge and A and B are equidistant from the center of the edge.
Can you prove symmetry for other cases?
And how do you fold or transform a 3D cube (or, if there is symmetry, a half of a 3D cube, or a quarter of a 3D cube) into a square (flat) lamina with isotropic properties?
Also, just curious: is this something with practical application?