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!

  • $\begingroup$ $V$ is the potential on the charged plate, not in free space. $\endgroup$ – probably_someone Jul 1 '19 at 11:57
  • $\begingroup$ Are you inferring that at each snapshot in time, the potential is uniform across the plate? Thus allowing $ E = \frac{V}{d} $ to be used? $\endgroup$ – Pox 219 Jul 1 '19 at 13:24
  • $\begingroup$ If your problem is 1-dimensional as you say, then the voltage has to be uniform across the plates, which also have to be large enough that they can be reasonably approximated as infinite. Any other configuration would introduce some 2-dimensional variation in the electric field. Whether or not $E=V/d$ can be used depends on if the voltage is varying slow enough that you can neglect the fact that voltage changes propagate at the speed of light. $\endgroup$ – probably_someone Jul 1 '19 at 13:33
  • $\begingroup$ I see. That makes sense, thanks for your help $\endgroup$ – Pox 219 Jul 1 '19 at 14:00

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.

  • $\begingroup$ If someone does have a solution to what the "relativistic" field inside a capacitor looks like, I would be more than happy to hear about it. $\endgroup$ – Andréas Sundström Jul 1 '19 at 16:25
  • $\begingroup$ Many thanks for that. I am writing some code so I did make sure the particle didn't reach the plates. I have one further question if that is ok. I'd like to plot conservation of energy. So E = KE+W. I obtain KE by plugging in $v$ from the equation above into $\frac{1}{2}mv^2$ . However when I plot this and W against time, W completely dominates KE, causing the total energy E to oscillate. Am I right in thinking that as $W=\int_{0}^{d} eE dx$, then $W = \frac{eV0cos(\omega t)}{d} \int_{0}^{d} dx$ ? $\endgroup$ – Pox 219 Jul 2 '19 at 14:44
  • $\begingroup$ The (again pretty short) answer is that since the potential (and thus also the force field) has an explicit time dependence, $V=V(z;t)$, energy is not conserved. If you solve the particle trajectory with the simplifying assumptions, you get $E_{tot}=...=\frac{1}{2m}\left(\frac{eV_0}{\omega d}\right)^2[\cos^2(\omega t)-2\sin(\omega t)]-eV_0z_0/d$ for an electron, which clearly isn't constant. If you want to look up why energy isn't conserved in time-independent force fields, have s look at Noether's theorem (but bewarned some pretty deep theory lies in those paths). $\endgroup$ – Andréas Sundström Jul 3 '19 at 15:26
  • $\begingroup$ Ooups! I forgot a factor $\sin(\omega t)$ on the last term in the expression for the total energy. The correct expression should be $E_{tot}=\frac{1}{2m}\left(\frac{eV_0}{\omega d}\right)^2[...]-eV_0z_0\sin(\omega t)/d$. But this gives me an opportunity to again demonstrate that enegy doesn' have to be conserved in a time varying force/potential field: let $m\to\infty$, then the particle would not move, but since the potential at that spot ($z_0$) is still varying, so must the total energy. $\endgroup$ – Andréas Sundström Jul 3 '19 at 15:47

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