Current Density and Conductivity and Electric Field

[Oijl]

In the time between posting this and now, I have found my mistake. So, problem solved.

Homework Statement

Earth's lower atmosphere contains negative and positive ions, created by radioactive elements in the soil and cosmic rays from space. In a certain region, the atmospheric electric field strength is 140 V/m and the field is directed vertically down. This field causes singly charged positive ions, at a density of 610 cm^-3, to drift downward and singly charged negative ions, at a density of 550 cm^-3, to drift upward. The measured conductivity of the air in that region is 2.70 multiplied by 10-14 (ohm·m)^-1.

Calculate the magnitude of the current density.

Homework Equations

J = conductivity * E
J = nev
(v here represents the drift speed)

The Attempt at a Solution

I was writing this and lost what I wrote (alas, I was too verbose!), so here's it quick:

J = conductivity * E = 2.7*10^-14 * 140 = 3.78*10^-12 <<------- correct
v = J / ne
Here we have two values for n, one of positive ions and one of negative ions. So I figured I could write:
v = J / (e(n1-n2))
where n1 represents the density of the positive ions and n2 of the negative.

I checked my math a whole lot, but I keep coming out with v = 0.393258427 m/s, which is wrong, says www.webassign.com.

So apparently I can't just take the net drift?

 

[Redbelly98]

That's weird, you're solution looks good to me.

 

[Oijl]

Ya, actually, it was correct, I was just putting it in wrong. (So I'm sorry for putting up a question for such a ridiculous reason.) (I'm not being sarcastic, since this is just text and that can be hard to tell sometimes.)

But anyway, the second part of the question is the real thing that's stumping me. And so I changed the first post.