Consider an electron deflection tube (something like this: http://www.ld-didactic.de/documents/en-US/GA/GA/5/555/555624e.pdf).

Suppose you apply a magnetic field perpendicular to the screen. Then the electron beam will be deflected. Say the orientation of the magnetic field is such that it will be deflected to the upper capacitor plate. The electrons hitting this plate will charge it up. I expect that after some time the charge will be large enough that the effects of the electric and magnetic fields will cancel so that the beam will follow a straight path. Same idea as in the hall effect.

How can I estimate quantitatively if this will work or not with the tube linked above?

Here is what I tried so far:

Plate distance: $d = 50\,\mathrm{mm}$
Plate Area: $A = 10\,\mathrm{cm} \cdot 5 \, \mathrm{cm} = 5\cdot 10^{-3} \,\mathrm{m}^2$

Thus the capacity is

$$ C = \epsilon_0 \frac{A}{d} \approx 8,85\cdot 10^{-13} \,\mathrm{F} $$

Assume an magnetic field of $3 \,\mathrm{mT}$ and an acceleration voltage of $U_A = 4\, \mathrm{kV}$, so we get a speed of $v = \sqrt{\frac{2U_Aq}{m}} \approx 3,77\cdot 10^7\,\mathrm{\frac{m}{s}}$

Balance of forces leads to:

$$ Bv = E = \frac{Q}{Cd} $$

And plugging in the numbers:

$$ Q = BvCd \approx 5\,\mathrm{nC} $$

Is this correct so far? Now I want to calculate the time it takes to charge the capacitor with this charge by the electron beam but I don't have a number for the current.

I tried this experiment in the lab (also plugged in an electroscope to the upper plate) but the electroscope did nothing.

  • $\begingroup$ Unfortunately most of the electrons will bounce off of the capacitor plate. The passing electron beam willinduce charge in the plate, a positive charge, which will slightly attract the beam. The energy of the electrons is a very important factor in these calculations. $\endgroup$ Jun 14, 2016 at 12:17
  • $\begingroup$ @PeterDiehr Why doesn't this happen in the case of the hall effect? Can you give some more details? $\endgroup$
    – Julia
    Jun 14, 2016 at 12:25
  • 2
    $\begingroup$ The charge carriers in a solid travel at the drift velocity. This is very, very slow when compared to an electron beam. Your beam is travelling millions of meters per second vs cm per second. I've actually done your experiment by accident - a metallic mirror near the electron beam would slowly charge up, deflecting the 20 kV electron beam slightly. As for bouncing off of flat plates, look up design of Faraday cups, used to collect electron beams. $\endgroup$ Jun 14, 2016 at 12:32
  • $\begingroup$ Isn't your current dependent on your electron gun? $\endgroup$
    – Melvin
    Apr 28, 2018 at 17:52

1 Answer 1


Even if every electron that hits the plate sticks to it, a 4 KeV electron beam cannot charge a 5 cm gap capacitor enough to cancel out a 3 mT magnetic field. Even more problematically, a high energy electron beam hitting the plate can knock out more electrons than it adds, which may limit the maximum possible charge on the plate.

The force on an electron with charge $e$ travelling with velocity $\vec{v}$ in an electric field $\vec{E}$ and magnetic field $\vec{B}$ is given by the Lorentz Force Law: $$\vec{F}=e(\vec{E}+\vec{v}\times \vec{B})$$ For the electrons to travel in a straight line perpendicular to $\vec{B}$, the electric and magnetic forces must cancel out, so $$E=-vB$$ As noted in the comments, the magnitude of the velocity makes a big difference. The drift velocity in copper might be only $2\times10^{-5}$m/s, so to cancel out a magnetic field of $B=3$mT only requires a tiny electric field of $E=60$nV/m. Electrons in your beam, however, have a kinetic energy of 4 KeV and a velocity of about $3.8\times10^{7}$m/s (ignoring the ~1% relativistic correction). This would require an electric field that is 12 orders of magnitude larger, i.e. 0.11 MV/m. For plates that are $d=5$ cm apart, this requires them to reach a potential of about $V_{plate} = 6$ KV.

Electrons accelerated by a 4 KV potential cannot charge the plate to 6 KV. The beam charges the plate negatively and once the potential of the plate reaches the accelerating potential of the beam, electrons can no longer reach the plate and the capacitor stops charging. In the non-relativistic limit (where kinetic energy = $m v^2/2$), the minimum accelerating voltage that allows electrons to reach the plate works out to be $$V_{acc}=\left(\frac{V_{plate}}{B d}\right)^2\frac{m_e}{2 e}$$ We must have $V_{acc}/V_{plate} > 1$ to charge the plate, which requires $$V_{acc} > 2 \frac{e}{m_e} B^2 d^2$$ This minimum voltage is about 8 KV for $B=3$mT and $d=5$cm, so let's assume you can apply a higher accelerating voltage, say 10 KV, that is sufficient. (The Leybold tube actually has a maximum safe anode voltage of 5 KV, but we will assume it is over-engineered and can support 10 KV.)

This still cannot charge the capacitor sufficiently to produce a straight beam, however, since as the electric field increases it bends the beam away from the plate so that it eventually misses the plate and the charging stops. To produce a perfectly straight beam you would need an infinitely long capacitor which would, of course, take infinitely long to charge.

But even an infinitely long capacitor is unlikely to work. We have been assuming that every electron that hits the plate sticks to it, but as noted in the comments, this isn't true. In fact, it is worse than this - the beam can actually cause the plate to lose electrons. If the beam is narrow and powerful, it may heat up a small area on the surface and electrons will boil off by thermionic emission. Even if the metal doesn't heat up, when a high energy electron hits a metal surface it can eject electrons through secondary emission.

The number of secondary electrons produced per incident electron is known as the secondary emission coefficient $\delta$. For the beam to charge a single plate, $\delta$ must be less than 1. $\delta$ varies with energy, typically with a peak greater than 1 for incident electrons with a few hundred electron-volts energy, and falling at higher energies because the incident electrons penetrate so deeply that excited electrons are too deep to escape easily. For electrons incident normal to a carefully prepared metal surface, $\delta < 1$ is possible at high enough energies, so charging a single plate might be possible, but $\delta$ increases with decreasing incidence angle since the electrons penetrate less deeply, which will eventually limit the maximum charge build-up. As the plate charges up, the incident electron energy is reduced by the repulsive electric forces, which increases $\delta$ and also limits the charge build-up.

For two isolated parallel plates, $\delta < 0.5$ is needed to charge the plates to any significant voltage because secondary electrons cause a current to flow between the two plates discharging the capacitor. The problem is that once any significant negative charge has built up on the plate hit by the beam, there will be an electric field between the plates so that when an a secondary electron is knocked out of the negative plate, it will likely travel to the other (relatively positive) plate and increase the negative charge on that plate, reducing the electric field between the plates. An incident electron hitting the negative plate charge will on average increase that plate's charge by $(1-\delta)e$, while the positive plate charge will increase by $\delta e$, so the charge difference between the plates will be $(1-2\delta)e$. For the charge difference between the plates to build up, we need $\delta<0.5$, which would be unusual.

It seems unlikely that the electron beam can significantly charge the capacitor.


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