Resistance of a odd shaped conductor

[jesuslovesu]

Homework Statement

http://img228.imageshack.us/img228/2536/physicsme1.th.png

Homework Equations

$$I = \int \int J . dS$$
$$V = - \int E . dl$$
$$V = IR$$

The Attempt at a Solution

I'm not really sure how to get started on this problem, generally I would start with Laplace's equation but due to its irregular shape, I don't think that I can use that for this situation.
J is a function the length of the shape, but not sure how to formulate it into something useful
Does anyone know how I can get started on this?

$$I(z) = \int_0^h \int_0^{2\pi} \mathbf{J}(z) z d\phi dz$$

 

[alphysicist]

Hi jesuslovesu,

Here are a couple of points that may help.

The resistance of an object is R=V/I. Here you can assume that there is a constant current coming in one side of the resistor and leaving the other (and they must be the same-do you see why?) So if you can find the potential difference across the resistor that gives that current you'll be done.

The resistivity relates the current density and the electric field by

$$\vec J = \frac{\vec E}{\rho}$$

You say you found an expression for $\vec J$. How can you then find V?

 

[jesuslovesu]

Hey thanks for your reply, I've almost got it but I can't seem to quite get the last part.

So I say that $$J(z) = \frac{I}{\pi r^2}$$
$$V = - \int_0^h \frac{I dz}{\sigma \pi r^2}$$
$$\frac{I*h/(\sigma \pi r^2)}{I} = \frac{h}{\sigma \pi r^2}$$
I've almost got it but I don't quite know what to do with the 'r', I am not positive if my integral is correct since J is a function of z

 

[berkeman]

I would have approached it differently, but either way should be valid. I would have started with the equation for the resistance of a fixed-area element:

$$R_i = \frac{\rho L}{A}$$

and integrated circular cross-sections of the conic section. Maybe give that approach a try...

 

[kamerling]

r is a function of z. You have r(0) = a and r(h) = b