# Resistance, Current, two cylinders

1. Oct 6, 2011

### JosephK

1. The problem statement, all variables and given/known data

An oceanographer is studying how the ion concentration in seawater depends on depth. She makes a measurement by lowering into the water a pair of concentric metallic cylinders at the end of a cable and taking data to determine the resistance between these electrodes as a function of depth. The water between the two cylinders forms a cylindrical shell of inner radius $r_a$, outer radius $r_b$, and length L much larger than $r_b$,. The scientist applies a potential difference ΔV between the inner and outer surfaces, producing an outward radial current I. Let ρ represent the resistivity of the water.

2. Relevant equations

ΔV = El
R = ρ l / a

3. The attempt at a solution

The potential difference is maintained across concentric metallic cylinders. We know for a wire of length l with a potential difference maintained across is ΔV = El. In this case, an outward radial current I is produced. In other words, current flows inbetween the cylinders. Thus, charges flow from $r_a$ to $r_b$. Hence, length l is $r_a$ - $r_b$. Finally, ΔV = E($r_a$ - $r_b$).

In the equation R = ρ l / A, A is cross-sectional area of the conductor (saltwater is good conductor) which is pi(b^2-a^2).

So, I learn that current density equation is only valid if the cross sectional area A is perpendicular to the current density. So, an area perpendicular to the current are cylinders.

So A = 2 pi r L, for some radius r.
Then, I need integrate from $r_a$ to $r_b$ with respect to r.

So, I write

dA = 2pi r L dr ?

The reasoning for length is still valid.

So, ∫dR = ∫$^{r_b}_{r_a}$ρ ( ($r_a$ - $r_b$) / (2pi r L dr)

R = $\frac{ρ(r_a - r_b) }{2\pi L }$∫$^{r_b}_{r_a}$ $\frac{1}{ r dr}$

R = $\frac{ρ(r_a - r_b) }{2\pi L }$ ln$\frac{r_a}{ r_b}$.

2. Oct 6, 2011

### Staff: Mentor

Your reasoning looks good up until you write the integral. Why do you include the (ra - rb) term in the integral? Does it have some physical significance to the differential resistance element?

3. Oct 7, 2011

### JosephK

I have trouble with integrals.

4. Oct 7, 2011

### Staff: Mentor

The integral is a sum of differential elements. In this case you're trying to compute a total resistance, so you should expect to be adding differential elements of resistance, or dR.

The resistivity formula you have for a slab of material is R = ρl/A, where l is the 'thickness' of the slab and A the cross sectional area perpendicular to l. You've identified A as being 2πrL for a given radius r. The thickness of a differential element is dr, and so that is identified with l. You should, then, be able to directly 'transcribe' the resistivity formula into a formula for dR.

5. Oct 7, 2011

### JosephK

I wrote formula for dR without recognizing dr.

6. Oct 7, 2011

### JosephK

(b)

$\rho = \frac{1}{\sigma}$

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

$\rho = \frac{AE}{I}$ where A = 2 $\pi$ rdr

$\rho = \frac{A \Delta V L}{I}$ where L = r

Does $\rho$ depend on geometry?
Why do I need to integrate?

7. Oct 8, 2011

### JosephK

ρ=AΔVL / I where L = r

should be

ρ=AΔV/rI

then from part a, replacing A and r,

ρ=(2pi L)ΔV/(ln(r_b / r_a) I