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Electron in an oscillating electric field

  1. Jul 7, 2013 #1
    1. The problem statement, all variables and given/known data
    Describe the motion of a free electron with charge -e when subject to an oscillating electric field along the x-axis: $$E_x = E_o \cos(\omega t + \theta)$$


    2. Relevant equations

    Force acting on the electron: $$F=-eE_x=-eE_o \cos(\omega t+ \theta)$$

    (*) Equation of motion: $$F = m \frac{dv}{dt} = -eE_o \cos(\omega t+ \theta)$$

    3. The attempt at a solution

    Multiply (*) by ##dt## and integrate taking ##t_0=0##.
    $$v = \frac{dx}{dt} = v_o + \frac{e E_o \sin(\theta)}{m \omega} - \frac{e E_o}{m \omega}\sin(\omega t +\theta)$$

    Multiply by ##dt## again and integrate. Taking ##v_0=0## and ##x_0=0##
    $$x = -\frac{e E_o \cos(\theta)}{m \omega^2} + \frac{e E_o \sin(\theta)}{m \omega} t + \frac{e E_o}{m \omega^2}\cos(\omega t +\theta)$$


    So I'm left with three terms in the final equation, and I'm not quite sure how to explain them all. Partly, I'm not sure that I'm imagining the electric field correctly.

    (1) The first term is a constant which appears to account for the initial phase of the electric field at ##t=0## therefore making sure ##x=0##.

    (2) The second term is linear in ##t## but I can't come up with a good explanation why. I assume that it accounts for the constant movement of the electron along the x axis, maybe due to the initial momentum imparted on the electron by the electric field (sort of an initial "drift"). Always positive or negative depending on the initial phase, θ.

    (3) The third term seems to be due to the oscillation of the electric field. It is of opposite sign of the electric field itself because of the negative charge on the electron.

    Does this sound correct? Can anyone help me understand the second term?

    Thanks guys
     
    Last edited: Jul 7, 2013
  2. jcsd
  3. Jul 7, 2013 #2

    TSny

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    I think your interpretations of the terms are good.

    To get some insight into the second term, you can take the special case of ##\theta = \pi/2##. Think about how the position and velocity of the particle change during the first half cycle of the driving force and then how they change during the second half cycle. You might want to draw a rough sketch showing the particle's location on the x-axis at t = 0, T/8, 2T/8, 3T/8,...2T (where T is the period of the driving force) and include vectors drawn roughly to scale showing force and velocity at those instants. Does the particle ever have a negative velocity during the full cycle? What is the velocity after one cycle? Think about how the inertia of the particle plays a role in what's happening.
     
  4. Jul 7, 2013 #3

    dlgoff

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  5. Jul 7, 2013 #4
    Ahhh yes it makes much more sense now that I've drawn it out. When t=T/4 (for ##\theta=\pi/2##), the velocity is at a maximum. This also corresponds to the maximum value of the third oscillatory term in the position equation; the the first term (which is zero) and second term are not effected. The velocity will never be negative so there has to be a term in the position equation that accounts for this continual +x movement, and that is the second, linear term.

    I appreciate your help!
     
  6. Jul 7, 2013 #5

    TSny

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    That's not quite correct. The maximum velocity does not occur at T/4.

    Yes.
     
  7. Jul 7, 2013 #6
    Are you sure? Hm... when t=T/4, $$\omega t=\frac{2 \pi}{T} \frac{T}{4}= \frac{\pi}{2}$$

    Therefore $$\sin(\omega t + \frac{\pi}{2})=\sin(\pi)=0$$

    $$v = \frac{e E_o }{m \omega} - \frac{e E_o}{m \omega}\sin(\omega t +\frac{\pi}{2}) = \frac{e E_o }{m \omega}$$

    and ##v## is at a maximum value, right?

    What am I missing?
     
  8. Jul 7, 2013 #7

    TSny

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    Try evaluating v at ##t = T/2##.

    Is the force positive or negative for ##T/4< t < T/2##? What does that tell you about what the velocity is going to do in that interval of time?

    Would you expect the velocity to have maximum value where the force is maximum?
     
  9. Jul 7, 2013 #8
    Oh yes, I appologize :redface:
    Of course the change in signs would make it positive so the max is $$\frac{2eE_0}{m\omega}$$

    The force would be positive and approaching zero. The velocity will be increasing in the +x direction.

    No, because the force will be causing the velocity to increase even when the force is transitioning from max value to 0. Only when it changes signs does the force begin to decrease the velocity.
     
  10. Jul 7, 2013 #9

    TSny

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    Yes. That all sounds very good.
     
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