Sometimes you want to have a beam of high velocity electrons. You can use a strong electric field to accelerate an electron to a high velocity.

The term electron gun makes most people think of James Bond style evil boss, scenarios where an evil madman is about to point his super powerful weapon on unexpecting cities of peaceful citizens.

Come to think of it, it is kind of like that in the real world actually. You see, TV was based on the electron gun and it is a super powerful weapon that has been unleashed on the citizens. TV has caused more damage to humanity (health, intellectual, political, etc.) than all of the worlds other media put together. Even today as the internet becomes the ultimate source of low quality informational content, nothing beat the CNN news on bullshit to content ratio. TV is magical in that it tells people what to think, what to feel and what to buy.

What I am trying to say is that the electron gun is part of the system. As concerned citizens of the XXI century we must know and understand the weapons of the system, if we are to defeat it. So let's see how this electron gun works.

• $q_e=-1.602\times10^{-19}$[C]: charge of an electron
• $V$: Potential difference of the gun. The bigger this number is, the faster the electron will be accelerated.
• $\vec{E}$: The electric field inside the electron gun
• $d$: The distance between the two acceleration plates. This is in some sense the length of the “barrel” of the gun.
• $v_{exit}$: The velocity of the electrons as they leave the gun
• $a$: The acceleration that the electron feels while inside the gun.
• $m=9.10938188\times10^{-31}$[Kg]: The mass of an electron

To build an electron gun you first need a really hight voltage. Think 3000 [V] and then connect that voltage across two metal plates separated by a distance $d$. Now you have yourself a nice capacitor like device (two nearby plates one with some positive charge, one with negative charge on it).

Now make a tiny hole in the positive plate: this is where the electrons will come out from. Assume for the moment that around the negative plate there are plenty of electrons that can be stripped from the plate (in practice you might have to heat the negative plate).

Let's see what happens to an electron that is released somewhere near the negative plate. We know that a charged particle moving through a voltage $3000$ [V] will gain a total electric potential energy of \[ U_e = qV = q 3000. \qquad \text{[J]} \] But this is an electron, so it must have lost $q 3000= 1.602\times10^{-19} \cdot 3000$ [J] of energy.

Where, oh where did that energy go? If you look closely, you will see that the electron is flying now at a high velocity and those that do not hit the positive plate will come out of the hole with a speed of $v_{exit}$.

Basically the potential energy that an electron $e$ with charge $q_e = -1.602\times10^{-19}$[C] lost has been converted into kinetic energy $KE=\frac{1}{2}m_e v_{exit}^2 $.

By the law of conservation of energy, we can write that \[ U_{e,i} = K_f \] \[ q V = \frac{1}{2}m_e v_{exit}^2, \] or solving for $v_{exit}$ we get \[ v_{exit} = \sqrt{ \frac{2q V}{ m_e } }. \]

If we now substitute numbers we would get \[ 1.602\times10^{-19} \cdot 3000 = \frac{1}{2}(9.10938188\times10^{-31} ) v_{exit}^2, \] \[ v_{exit} = \sqrt{ \frac{2(1.602\times10^{-19})(3000)}{ 9.10938188\times10^{-31} } } = 3.24 \times 10^7 \qquad \text{[m/s]}. \]

check

In the above story we used the energy picture and the calculation was quite simple. Suppose that we wanted to do this with forces.

$\vec{F}_e = q \vec{E}$

what is the electric field in a between two plates?

\[ \vec{E} = \frac{V}{d} \]

So the force is \[ \vec{F}_e = q \frac{V}{d} \] and it is constant.

Now we know \[ \vec{F} = m \vec{a}, \] so the acceleration of the electron during the whole time it is flying between the two plates is going to be \[ \vec{a} = \frac{ \vec{F} }{ m } = \frac{ q \frac{V}{d} } { m } = \frac{ q V} { m d } \]

Now we need to remember the usual laws of kinematics. What will be the final speed of a particle if it is starting from a \[ v^2_f = v_i^2 + 2ad \] In our case we start from $v_i=0$ and $v_f = v_{exit}$ we we get \[ v_{exit}^2 = 2 \frac{ q V} { m d } d = \frac{ 2 q V} { m } \] and \[ v_{exit} = \sqrt{ \frac{ 2 q V}{ m } }. \] Which is good. Physics should be consistent.

see also:
wps.aw.com/wps/media/objects/877/898586/topics/topic07.pdf