1. Nov 4, 2004

### stunner5000pt

There is a spherical resistor witn an inner radius a and outer radius of b. The outer surface and the inner surface are covered with conducting sheets. Find the resistance betwen the two surfaces assuming uniform resistivity $$\rho$$

WELL

I know $$R = \rho \frac{L}{A}$$
If i divided the sphere into many cylinders then each cylinder would have a length of (b-a) and then the area would be ab times the radius (b-a)?

so then i end up with $$R = \rho \frac{1}{(b-a)ab}$$

but that isnt right because i know the answer has someting to do with 4pi? plase help!

Last edited: Nov 4, 2004
2. Nov 4, 2004

### maverick280857

This is an example of a continuous distribution of resistive material. That should give you a hint of using calculus. So instead of R = rho * L/A you will get

$$dR = \rho\frac{dL}{A}$$

and proceed.

3. Nov 4, 2004

### stunner5000pt

but what is the expression for A? area of this box would be (b-a) times ....??

4. Nov 5, 2004

### Tide

From first principles the current density j (current per unit area) is

$$j = \sigma E = -\frac {1}{\rho} \frac {\delta V}{\delta r}$$

where $\sigma = 1/\rho$ is the conductivity. Now

$$j = \frac {I}{4\pi r^2}$$

for your problem and it follows that

$$\Delta V = -\int_{a}^{b} \frac {\rho I}{4 \pi r^2} \delta r$$

or

$$\Delta V = \frac {\rho I}{4 \pi} \left( \frac {1}{a} - \frac {1}{b}\right)$$

The resistance is just the right hand side divided by I. Notice that in your original expression the units are not correct which is a hint that something was amiss.

5. Nov 5, 2004

### stunner5000pt

thank you very much!
that wasn't an approach i was thinking about but it works , and how!

thank you very much again!

6. Nov 5, 2004

### maverick280857

The dR approach is equivalent to Tide's approach. When I gave you the hint to use calculus, you should've realized that I am referring to the area of a spherical surface which would be 4pi (r squared).