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auk411

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Help much needed! Thanks!

So this problem has been dealt with here on the physics forums. But the fact is that answers either appear to be false and/or they differ from what other people have done.

First, here's the problem:

In a certain region, the atmospheric electric field strength is 120 V/m and the field is directed vertically downward. This field causes singly charged positive ions, at a density of 620cm^-3, to a drift downward and singly charged negative ions, at a density of 550cm^-3, to drift upward. The measured conductivity of air in that region is 2.70x10^-14 (ohm*m)^-1. Calculate the magnitude of the current density and ion drift speed. Assume it is the same for positive and negative ions.

[1] Here's the first attempt:

v = J / (e(n1-n2))

J = conductivity * E = 2.7*10^-14 * 120 = 3.24*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.

v = 0.393258427 m/s

[2] Second attempt:

J=nevd = (n+ n-)*e*vd // this time we're adding n.

3.24*10^-12 = (620-550)*10^-6*1.602*10^-19*vd

v=0.289m/s.

[3] Third attempt:

(n_total)*e*vd = J // adding n again.

v= 3.24x10^-12/ [(6.2+5.5)*10^8 (1.6x10^-19)]

v =1.73cm/s

So what is v (the velocity of the drift)? Some people add the n's. Some subtract. Which one to use? Other times, I can't tell if my algebra is off (other theirs). Cause I can't get their answer.

Also, how do we know when to add n

So this problem has been dealt with here on the physics forums. But the fact is that answers either appear to be false and/or they differ from what other people have done.

First, here's the problem:

In a certain region, the atmospheric electric field strength is 120 V/m and the field is directed vertically downward. This field causes singly charged positive ions, at a density of 620cm^-3, to a drift downward and singly charged negative ions, at a density of 550cm^-3, to drift upward. The measured conductivity of air in that region is 2.70x10^-14 (ohm*m)^-1. Calculate the magnitude of the current density and ion drift speed. Assume it is the same for positive and negative ions.

## The Attempt at a Solution

[1] Here's the first attempt:

v = J / (e(n1-n2))

J = conductivity * E = 2.7*10^-14 * 120 = 3.24*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.

v = 0.393258427 m/s

[2] Second attempt:

J=nevd = (n+ n-)*e*vd // this time we're adding n.

3.24*10^-12 = (620-550)*10^-6*1.602*10^-19*vd

v=0.289m/s.

[3] Third attempt:

(n_total)*e*vd = J // adding n again.

v= 3.24x10^-12/ [(6.2+5.5)*10^8 (1.6x10^-19)]

v =1.73cm/s

So what is v (the velocity of the drift)? Some people add the n's. Some subtract. Which one to use? Other times, I can't tell if my algebra is off (other theirs). Cause I can't get their answer.

Also, how do we know when to add n

_{-}and n_{+}) and when to subtract. (n is the number of carriers per unit volume).
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