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Unknown008 · Apr 23, 2010, 11:53 PM

I had a question in my physics book which I can't solve... I don't even know where to start :(

The ionosphere contains free electrons. What is the amplitude of these electrons when subjected to a 200 kHz electromagnetic wave in which the oscillations of electric field have amplitude 5 x 10^-3 V/m?

A. 3.2 x 10^-15 m
B. 4.0 x 10^-9 m
C. 2.5 x 10^-8 m
D. 5.6 x 10^-4 m
E. 2.2 x 10^-2 m

~~~~
Electrons have a charge of 1.6 x 10^-19 C... So, force is 8 x 10^-22 N. I don't even know if that force will be needed :(

I tried using
$x = x_o sin (\omega t)$

but not only I get a negative number, but the number lies between the answers of D and E.

Alty · Apr 24, 2010, 12:12 AM

Okay, Unky, all I can say is this. If you can't solve it, I have no hope in hell of even getting close.

I hope someone can help. I know it's not me.

Good luck kiddo. :)

KISS · Apr 24, 2010, 12:44 AM

Would it come from E=kQ/R^2
k = permetivity (see Permittivity - Wikipedia, the free encyclopedia
Q = charge on electron
solve for 2R

Just an idea.

Unknown008 · Apr 25, 2010, 01:44 AM

Ok, so you're probably suggesting this then:

$E = \frac{Q}{4\pi \epsilon _o r^2}$

?

I know this formula, and epsilon nought is the permittivity of free space.

The problem is, I can't find the link between the question and the solution. I know I can use that formula in cases for example a charged sphere, and then, I find the electric field at distance 'r' from the sphere. Here, there is no 'charge centre', and I'm at a loss... =/

KISS · Apr 25, 2010, 10:11 AM

K varies with f. Find k.
See link.

Unknown008 · Apr 25, 2010, 10:36 AM

Um.. still confused. K is a constant...

If you're telling me to find the force... wait, there aren't any information to find that. I need to get either Q or r to get that, and that charge is not due to the electron but to the source of the electromagnetic wave...

Ebaines, some input from you would be welcome too!

KISS · Apr 25, 2010, 11:52 AM

From the link, I woas sort of thinking of this:

instantaneously to an applied field. The response must always be causal (arising after the applied field) which can be represented by a phase difference. For this reason permittivity is often treated as a complex function (since complex numbers allow specification of magnitude and phase) of the (angular) frequency of the applied field ω, $\varepsilon \rightarrow \widehat{\varepsilon}(\omega)$ . The definition of permittivity therefore becomes

$D_0 e^{-i \omega t} = \widehat{\varepsilon}(\omega) E_0 e^{-i \omega t}$ ?

where

D0 and E0 are the amplitudes of the displacement and electrical fields, respectively,
i is the imaginary unit, i 2 = −1.

Problem is, what is $\widehat{\varepsilon}(\omega)$

Unknown008 · Apr 25, 2010, 12:06 PM

Uh... I don't think my syllabus requires me to go that far... :eek:

InfoJunkie4Life · Apr 25, 2010, 04:45 PM

1 N/C = 1 V/m Helpful?

I know I can't solve it, but I am going to speculate on some problems I see with this problem.

The amplitude of the radio frequencies is in V/m while the ionosphere is only given in m (answers).

If you're talking force, wouldn't the electron density of the ionosphere be helpful.

Some other things I know that aren't mentioned. 250kHz is a strange frequency. Its right on the border between VHF and UHF. This makes it very susceptible to density. The electro-density of the ionosphere would determine weather it is bounced off, absorbed, or let through. There is nothing mentioned about the inverse square law.

I really don't have much of a clue, but these are a few things that popped into my mind as I read through the thread.

Good Luck.

Unknown008 · Apr 26, 2010, 06:06 AM

Thanks for your input. Yes, I knew that 1 N/C = 1 V/m, but I can't see where the electric field can be connected to the amplitude and the frequency.

I guess I'll have to wait until the day my teacher's going to give the correction for this number...

KISS · Apr 26, 2010, 06:03 PM

If I get a chance, I'll check my "Field theory" book if I can find it.

Unknown008 · Apr 26, 2010, 10:56 PM

Sure, thanks KISS :)

By Thursday, I'll be having the correction and I'll post it here, just in case you wanted to know :)

Unknown008 · May 8, 2010, 09:00 PM

Sorry for the late reply, I got the solution, then did a wrong calculation and thought that I missed something in the solution, but it was a mere calculation mistake. :o I really need to revise this chapter because the answer is in fact really easy...

Ok, here we go, KISS, you were nearer with your first idea ;)

Given the electric field, we can find the force that the electron experiences using:

$F = Eq$

Then, using the equation in oscillations, or circular motion, we know that:

$F = ma = m(r\omega ^2) = m \hat{A} $2\pi f$^2$

where r is the amplitude here.

Equating both, we get:

$m \hat{A} $2\pi f$^2 = Eq$

$\hat{A} = \frac{Eq}{m $2\pi f$^2} = \frac{(5\times 10^{-3})(1.6 \times 10^{-19})}{(9.11 \times10^{-31})(2\pi (200\ 000))^2}= 5.56 \times 10^{-4} = 5.6 \times 10^{-4}$

KISS · May 10, 2010, 02:48 PM

Seems simple, NOW!

Thanks.