Focusing a beam of electrons with a charged cylinder

First note that the tangential velocity of all the electrons are the same, since the spread angle is neglectble. Now note that the position of the electrons is a position of stable equilibrium, to prove this we can check the second derivation of the potential and note that it will be a negative term, we will now see this:

First note that the force is already the first derivation of the potential, deriving the force will give us the term we're looking for

F = pq/(2pier) - mw^2r^2

Where
q = electron charge p = linear density of charge e = electrical permissivity r = distance of the electron from the cilinder

We can now conservate the angular momentum, since all the forces are radial

L = m.w 0.r 0^2 = m.w.r^2 Then w^2 = w 0^2.(r 0/r)^4

Substituting this on the force equation we get:

F = pq/(2pier) - m(w 0^2.r 0^4).(1/r^3)

Derivating and aplying it on the equilibrium position we get

dF/dr = -p/(2pier^2) + 3m(w 0^2.r 0^4).(1/r^4)

But, on the equilibrium we have that

(w 0.r 0)^2 = pq/(2pim)

Where m = electron mass

Now, substituting all the values we get that our term ( we can call it M) can be written as:

M = - 2.m.w_0^2 < 0

It is proven that this is a stable equilibrium, wich means that the deflected electrons will expirience a MHS, the angular frequency of this movement can be easily calculated:

w^2 = -M/m = 2w_0^2

That means the deflected electrons will come to a maximum displacement and then return to the original trajectory, this will take a time t that is half the period of this MHS, using a simple rule:

2pi/x = T 0/t Where T 0 us the period of the rotation of the electrons around the cilinder and x is the angle we want

From our equations we get that T_0/t = 2.sqrt(2)

Then

x = pi.sqrt(2)/2 (in radians)

x = 127.28 (in degrees)

The force acting on the electrons can be found from Gauss' Law. If $\lambda$ is the cylinder's charge per unit length (height) then the electric field is given by $E=\frac{\lambda}{2\pi \epsilon_{0} r}$ where r is the distance to the center of the cylinder. Since the force is central, it is convenient to work in polar coordinates. In polar coordinates, Newton's second Law assumes the form $F_{r}=e E= ma= m (\frac{d^{2}r}{dt^{2}}-r( \frac{d\theta}{dt})^{2}).$ Note that the acceleration contains two terms. The first one corresponds to the radial acceleration while the second is just the centripetal acceleration. Let us introduce the constant $C:= \frac{\lambda e}{2\pi \epsilon_{0}m},$ and rewrite Newton's Law as $-\frac{C}{r}= \frac{d^{2}r}{dt^{2}}-r( \frac{d\theta}{dt})^{2}.$ For the electrons moving in circular orbits (i.e. $\frac{d^{2}r}{dt^{2}}=0$ ) we have $\frac{C}{r_{0}}= r_{0} \omega_{0}^{2}.$ Since the particles move under a central force, their angular momentum is conserved. In polar coordinates the angular momentum is $L= m r v_{\theta}= m r^{2} \frac{d\theta}{dt}.$ Now, since the electrons have equal initial speeds, say $v_{0}$ , their angular momenta are $L= m r v_{0} \cos(\alpha) \approx m r v_{0}.$ In other words, since the angle of divergence is small, all the electrons have the same angular momentum, regardless of their orbits. In addition, since we expect the deviations from the circular orbits to be small we write $r= r_{0} +\Delta{r}$ where $\frac{\Delta r}{r_{0}}\ll 1$ and $(1+ \frac{\Delta r}{r_{0}})^{n} \approx 1 + n \frac{\Delta r}{r_{0}}.$ Thus from Newton's Law and conservation of angular momentum we obtain (after simple calculations) the equation $\frac{d^{2} \Delta r}{ d t^{2}}+ 2 \omega_{0}^{2} \Delta r=0 .$ This equation should look quite familiar. It describes simple harmonic motion with frequency $\omega=\sqrt{2} \omega_{0}$ . Thus, we've learned that the electron's orbits are $r_{\alpha}(t) = r_{0} + A(\alpha) \sin( \sqrt{2} \omega_{0} t).$ where the amplitude $A( \alpha)$ depends on the initial direction of the electrons (how would you determine $A(\alpha)?$ ). The above equation is telling us something very important: at time $t =\tau:=\frac{\pi}{\sqrt{2} \omega_{0}}$ the beam focuses at point B. The angular displacement can be found as follows: $\Theta= \omega_{0} \tau= \frac{\pi}{\sqrt{2}}= 127.3^{ \circ}.$ The last step is not quite obvious and requires some clarification. From conservation of angular momentum we have $\frac{d \theta}{dt}= \omega_{0} \frac{r_{0}^{2}}{(r_{0}+\Delta r)^{2}}$ and therefore the angular speed of the electrons depends on their orbits. The angular displacement should be computed as follows: $\Theta= \omega_{0} \tau -2 \frac{\omega_{0}}{r_{0}}\int_{0}^{\tau} \Delta r(t) dt$ which is $\approx \frac{\pi}{\sqrt{2}}$ if the second term, proportional to $\frac{A(\alpha)}{r_{0}}\ll 1$ , is neglected. This approximation makes perfect sense!! Will the beam focus if we replace the cylinder with a charged sphere? What would change in that case? Note that our approach can be applied to any central force, in particular, it can be used to study the stability of planetary orbits.