Electric field of electron in alternating potential I am considering the case of an electron placed in an alternating sinusoidal potential, of frequency 13.56 MHz (frequency typically used for RF plasma generation). 
Suppose I have a simple 1D set up, with a charged plate and a grounded plate separated by $d$. 
Let $$ V = V_0 \sin(\omega t) $$ 
From here, I want to calculate $E$. Here lies my first problem. Generally, $E = -\nabla V$ . However, there is no space component in $V$, so I am unsure how to find $E$. Therefore, I state (I feel this stage is incorrect, since the field is not uniform but oscillating), that $$E = \frac{V}{d}$$
Since $F = m_ea = eE = e\frac{V}{d}$, I obtain,
$$a(t) = V_0 \sin(\omega t) \frac{e}{md}$$
Integrating and setting $v_0 = 0$,  we get,
$$ v (t) =  \frac{V_0e}{\omega md}(1-\cos(\omega t)   $$
These equations seem to make sense. If $d$ = 1m and $V_0$ = 1V, then the velocity would not exceed the speed of light. However I am just struggling to accept that $E = V/d$. I thought this was only the case for a uniform electric field. If I am correct in my doubt, please could I have some advice on how to find $E$?
Many thanks!
 A: Short answer
You have a couple of things to worry about:

*

*Will the particle hit the plates?
This, might be a little outside your question, but just keep that in mind.

*Do we have to worry about the finite propagation speed of the electromagnetic field?
As long as $\omega d\ll c$, where $c$ is the speed of light, we won't. Then you can safely assume that $E(t)=-V(t)/d$ homogeneously between your capacitor plats.
(The minus sign comes in if you say that the charged plate lies in the positive $z$ direction relative to the ground plate.)

*Do we have to worry about relativistic effects?
As long as the capacitor voltage is small enough such that $V_0q/m\ll c^2$, we an safely use Newtonian mechanics to solve for the particle dynamics.

If all of these conditions are fulfilled, then your solution is perfectly fine (modulo the minus sign depending on how you want to define positive direction).
Further elaboration
The relation $E=-V/d$ stems from electrostatics and is only valid when you can ignore the displacement current in Amprère's Law. You get to that solution by saying that there has to be some electric field, perpendicular to the plates, between the plates, but
$\nabla\cdot\mathbf{E}=\rho/\epsilon_0=0$
forces that field to be homogeneous. Hence $E$ is constant and the potential has to increase linearly from the ground plate (at $z=0$) to the other plate (at $z=d$): $V(x)=V_0 \frac{z}{d}$, $0\le z\le d$, which then gives you the electric field: $\mathbf{E}=-\nabla V(z)=-V_0/d\hat{z}$. This is a fairly classical electrostatics problem, see this question for more details.
For the full picture, you would need to solve Maxwell's equations including the displacement current:
$$
\nabla\cdot\mathbf{E}=\rho/\epsilon_0,\quad
\nabla\times\mathbf{E}=-\frac{\partial\mathbf{B}}{\partial t}
\\
\nabla\cdot\mathbf{B}=0,\quad
\nabla\times\mathbf{B}=\mu_0\mathbf{j}+\mu_0\epsilon_0\frac{\partial\mathbf{E}}{\partial t}
$$
This is a tricky endeavor which I unfortunately can't do for you off the top of my head. For instance, you can't use the friendly and cosy plane wave solution here, since the waves here are longitudinal. I would imagine that this would involve a non-spatially uniform field  and possibly also magnetic fields, which would require 2D or 3D in order to satisfy $\nabla\cdot\mathbf{E}=0$ (assuming that you haven't ionized your plasma yet).
As for the relativistic particle dynamics, all you have to do is replace $ma(t)$ with $\dot{p}(t)$, and then $mv(t)=p/\gamma=p/\sqrt{1+p^2/(mc)^2}$.
I hope this at least help you a bit on the way.
