Electron beam deflection relates to the behavior of a beam of electrons when fired from an electron "gun" and through a defined field. The diagram accompanying shows the path of a beam through an electric field, E, set up inside a cathode ray tube. Note that although cathode ray tubes are essentially extinct now, they still serve as excellent props for use in teaching electron beam deflection.

Since the E-field is vertical (+ to -) as shown in the diagram, no horizontal force act on the electron entering a region between the charged plates. Thus, the horizontal velocity component remains unaffected.

The displacement y in the vertical direction can be obtained from:

y = ½ at^2

Then, by Newtons' second law of motion (resultant force F = mass times acceleration):

m(e)a = Ee

where m(e) is the mass of the electron, E is the electric field intensity in V/m and e is the unit of electronic charge (e = 1.6 x 10^-19 C).

So the acceleration:

a = eE / m(e)

Therefore, the vertical displacement can be written:

y = ½ {eE/ m(e)} t^2

Meanwhile, horizontally:

x = vt so t = x/v

Therefore we may write:

y = ½ (eE/m(e)) x^2/v^2 = (eE/ 2 m(e)v^2) x^2

which we can easily see is of the algebraic form:

y = kx^2 (parabola)

A special condition obtains when the electron just passes the plates (at distance x = D) so the value of y there is:

y = eE D^2/ 2 m(e)v^2

Then the time for transit between the plates, t is:

t = D/v

and the horizontal component of the velocity is:

v_y = a(y) t = (eE/m(e)) D/v

Then, the angle at which the beam emerges from its horizontal path can be found:

tan Θ = v_y/v = (eE D/ m(e)v) 1/v = eE D/ m(e)v^2

*Problem for energized and interested readers*:

A beam of electrons moving with v = 1.0 x 10^7 m/s enters midway between two horizontal plates in a direction parallel to the plates which are 5 cm long and 2 cm apart, and have a potential difference V between them. Find V, if the beam is deflected so that it just grazes the bottom plate. (Take the electron charge to mass raito: e/m(e) = 1.8 x 10^11 C/kg).

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