# Magnitude of current passing through unit area in atmosphere(He2+ and O2-)

1. Jul 31, 2012

### dawn_pingpong

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

At a high point in the Earth’s atmosphere, He2+ ions in a concentration of 2.8 x 10^12 m^-3 are moving due north at a speed of 2.0x10^6 m/s. Also, a 7.0 x 10^11m^-3 concentration of (O2)- (dioxygen -)ions is moving due south at a speed of 7.2 x 10^6m/s. Determine the magnitude and direction of the net current passing through unit area.

2. Relevant equations
I have no idea how to even approach this question... I think there is some law or sth that I don't know.

3. The attempt at a solution

2. Jul 31, 2012

### voko

What do you think "current" is?

3. Jul 31, 2012

### dawn_pingpong

I=Qt (or V/R which is not very useful here...)

But I don't know how to apply the formula/other formulas. The answer is in terms of A/m2, so it is electric current.

4. Jul 31, 2012

### voko

Is it Qt or Q/t?

5. Jul 31, 2012

### dawn_pingpong

Sorry sorry my bad:( it's charge per unit time...

6. Jul 31, 2012

### dawn_pingpong

Urgh I have this junior olympiad tomorrow really last minute cramping of formulas:(

7. Jul 31, 2012

### voko

Q/t it is. Now, what do you think the direction of the net current is?

8. Jul 31, 2012

### dawn_pingpong

I think, it is moving to the north. the He has a higher concentration, and furthermore it's 2+. then Q of He per unit area is 1.6x10^-19 x 2 x 2.8 x 10^12=8.96 x 10^-7C
Similarly QO2=-1.12 x 10^-7C

Then how does speed and time come into play? Or... is it like add both together, and find the net (positive charge?) which will be 7.84x10^-7C

or is there something to do with KE=1/2 MV^2=pd X Q? I think can find m, v just take as velocity of helium, then find pd. (though I think this is going no where).

What is the relationship with time then?

9. Jul 31, 2012

### voko

Correct, the currents of oppositely charged particles flow in opposite directions, thus the net current flows where the positive particles flow.

Why is this "per unit area"? It is not. This is simply the concentration of charge per unit volume (cubic meter).

Same remark. Plus, think about the sign in connection to what you just figured out with regard to direction.

You have something that you know concentration of per unit volume. You know its speed. You need something per unit area per second. Think about the dimensions and then think whether you can give a plausible physical interpretation to the result of your thinking.

10. Jul 31, 2012

### dawn_pingpong

is it just Qv? (1/m^3 x m/s = 1/m^2s) then it's 7.84x10^-7 x 2.0 x 10^6 = 1.57A/m^2. But this is wrong. I have a feeling I missed out something else. (anyway it's not the correct answer...) the unit is correct though. So i suppose is manupilation of some of the numbers. ahh you are so deep...

okay I think it's the separate speed. So don't take the net charge, rather Q x v(He) + QV(O2). I think it's addition because like one is negative and other one is positive, and it is not counter productive (which I believed at first) And adding together gives 2.6C/M^2s= 2.6A/m^2 which is the Answer:D

I think I'm just manupilating the numbers to get the desired answer, and don't understand the true concepts... Searched online and found it's actually charge density. But I still don't get why it is related to velocity...

Thank you:D:tongue:

11. Jul 31, 2012

### voko

That's why I said you should try and interpret that physically. Q is charge density, i.e., charge divided by volume. You need to find the total charge passing per area per time. So you need to find some volume to multiply that charge with, then divide by area and time. That volume, physically, must contain all the charge that passes through the area during the time. What volume is that?

Exactly.