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jbrussell93
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Homework Statement
Describe the motion of a free electron with charge -e when subject to an oscillating electric field along the x-axis: $$E_x = E_o \cos(\omega t + \theta)$$
Homework Equations
Force acting on the electron: $$F=-eE_x=-eE_o \cos(\omega t+ \theta)$$
(*) Equation of motion: $$F = m \frac{dv}{dt} = -eE_o \cos(\omega t+ \theta)$$
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
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