In Fig. 28-31, an electron accelerated from rest through potential difference V
ID: 1987969 • Letter: I
Question
In Fig. 28-31, an electron accelerated from rest through potential difference V1=1.24 kV enters the gap between two parallel plates having separation d = 18.9 mm and potential difference V2= 63.6 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?
Chapter 28, Problem 9Explanation / Answer
Given that n electron accelerated from rest through potential difference V1=1.46 kV
the potential difference V_2 = 126.0 V
seperation between the two parallel plates is d = 28.4 mm
For the electron to travel a straight line the net force must be zero and the electric force is eaual to magnetic force
q E = q v B
B = E / v ...............(1)
The potential difference V_2 = E d
E = 126.0 V / 28.4*10^-3 m
= 4.437*10^3 V /m
1/2 mv^2 = q V_1
1/2 * 9.11*10^-31 kg ( v^2) = 1.6*10^-19 C * 1.46*10^3 V
speed of the electron is v = 2.26*10^7 m/s
substituting in the equation 1 we get
B = 4.437*10^3 V /m / 2.26*10^7 m/s
= 1.959*10^-4 T
In unit vector notation B = - (0.196 mT) k