In the figure, an electron accelerated from rest through potential difference V1
ID: 2195487 • Letter: I
Question
In the figure, an electron accelerated from rest through potential difference V1= 0.961 kV enters the gap between two parallel plates having separation d = 17.3 mm and potential difference V2= 60.4 V. The lower plate is at the lower potential. Neglect fringing 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
Electron energy E1 = 0.879 keV = 879*1.6022E-19 J Velocity v1 = sqrt(2E1/me) Felec = EQ (E is the electric field) Fmag = Qv1 X B; assuming v1 and B are perpencicular, F = Qv1B Thus EQ = Qv1B ==> B = E/v1 E = V2/d B = E/v1 = V2/(v1d) = V2/(d*sqrt(2E1/me)) = 138/(0.0184*sqrt(2*879*1.6022E-19 / 9.10938215E-31)) Is