Equations of motion of an electron emitted from a surface

Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
33 replies · 4K views
haruspex said:
What I wrote gives you a way to relate the potential of the disc to its charge density.
I am sorry, I just don't see it (get it).
 
Physics news on Phys.org
NB76 said:
I am sorry, I just don't see it (get it).
##V(x)=2\pi \sigma k(\sqrt{R^2+x^2}-x)\approx 2\pi \sigma kR(1-\frac xR)=2\frac{Qk}R(1-\frac xR)=V_0(1-\frac xR)##.
Equating that to the large x approximation, ##\frac{Qk}x##, gives a quadratic for estimating where you should switch from one approximation to the other.
 
The equations of motion for an electron emitted with an initial kinetic energy \(KE\) at an angle \(\alpha\) to the surface can be derived from the laws of conservation of energy and momentum. Since the electron is emitted with an initial speed \(v_0\), its initial kinetic energy is \(KE = \frac{1}{2} m v_0^2\), where \(m\) is the mass of the electron. Under the influence of the electric field created by the potential \(V\), the electron will change its speed and direction of movement. Taking into account the law of conservation of energy (\(E = KE + eV\), where \(e\) is the charge of the electron) and the law of conservation of momentum (\(mv_0 = mv + meV\)), we can obtain equations describing the trajectory of the electron.
 
What is the meaning of the meV term in the "conservation of momentum"? Is this a quantity with dimensions of momentum?