- #1
JosephK
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Homework Statement
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 [itex]r_a[/itex], outer radius [itex]r_b[/itex], and length L much larger than [itex]r_b[/itex],. 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.
Homework Equations
ΔV = El
R = ρ l / a
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 [itex]r_a[/itex] to [itex]r_b[/itex]. Hence, length l is [itex]r_a[/itex] - [itex]r_b[/itex]. Finally, ΔV = E([itex]r_a[/itex] - [itex]r_b[/itex]).
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 [itex]r_a[/itex] to [itex]r_b[/itex] with respect to r.
So, I write
dA = 2pi r L dr ?
The reasoning for length is still valid.
So, ∫dR = ∫[itex]^{r_b}_{r_a}[/itex]ρ ( ([itex]r_a[/itex] - [itex]r_b[/itex]) / (2pi r L dr)
R = [itex]\frac{ρ(r_a - r_b) }{2\pi L }[/itex]∫[itex]^{r_b}_{r_a}[/itex] [itex]\frac{1}{ r dr}[/itex]
R = [itex]\frac{ρ(r_a - r_b) }{2\pi L }[/itex] ln[itex]\frac{r_a}{ r_b}[/itex].
Please help