In the figure, an electron accelerated from rest through potential difference V_
ID: 1600291 • Letter: I
Question
In the figure, an electron accelerated from rest through potential difference V_1 = 1.00 kV enters the gap between two parallel plates having separation d = 20.0 mm and potential difference V_2 = 100 V. The lower plate is at the lower potential. Neglect fringing and gravity, and assume that the electron's velocity vector is perpendicular to the electric field vector between the plates. In unit-vector notation, what uniform magnetic field allows the electron to travel in a straight line in the gap?Explanation / Answer
Using energy conservation on electron before entering the plates
½ mv2 = eV
v2 = 2eV/m = 2*1.6*10-19 *1000/(9.1*10-31 ) = 3.5*1014
v = 1.87*107 i m/s
After entering in between plates , the force on electron due to electric field is given by
F = qE = qV/d = 1.6*10-19 * 100 / 0.02 = 8*10-16 j N
to balance this force we apply a magnetic field which will produce a force of 8*10-16 N in negative y-direction
using relation for magnetic force on a moving charge F = q(V X B)
-8*10-16 j= -1.6*10-19 * (1.87*107 i X B)
since i X k = -j hence B should be in -k direction
B = -8*10-16 /(1.6*10-19 * 1.87*107 ) k = -2.67*10-4 k T