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.

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. :)

Would it come from E=kQ/R^2
k = permetivity (see Permittivity - Wikipedia, the free encyclopedia (http://en.wikipedia.org/wiki/Permittivity_)
Q = charge on electron
solve for 2R

Just an idea.

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... =/

K varies with f. Find k.
See link.

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!

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)

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

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.

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...

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

Sure, thanks KISS :)

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

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}

Seems simple, NOW!

Thanks.