## Thursday, April 7, 2011

### Basic Electrodynamics- Solutions

We now revisit the solutions to the electron deflection problem at the end of the last instalment on Basic Electrodynamics. The problem was, again:

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).

The key piece of information (and key to the solution!) is that the beam just grazes the bottom plate. Since the total y -span = 2 cm, this means the effective beam deflection is one half this, or dy = 1 cm = 0.01m. Meanwhile, the electric field is:

E = V/d = V/ (0.02m)

The downward acceleration, from Newton's 2nd law, is:
a(y) = eE/m(e)

The vertical distance in terms of this acceleration is:

y = ½ {a(y) t^2) = ½ {eE/ m(e)} t^2

The time applicable is:

t = x/v = (0.005m)/ 10^7 m/s = 5 x 10^-9 s

and we know:

y = ½ (V/d)(e/m(e))t^2

Substituting:

0.01m = 0.5(V/0.02m) [1.8 x 10^11 C/kg] (5 x 10^-9 s)^2

for which V can be found:

V = 89 V