Question

a) An electron starts at rest and accelerates through an electric field
established by a set of parallel plates with a potential difference of 35 V.
What is the speed of the electron the instant before it hits the negative plate?
(e = 1.6 × 10^-19 C, melectron = 9.1 × 10^-31 kg)
b) Instead of hitting the negative plate, the electron, travelling East, escapes the
parallel plates through a small hole and enters a magnetic field of 0.75 T
directed downward. What will be the magnetic force (magnitude and
direction) on the charge?
c) Once the electron has entered the magnetic field, it is in circular motion.
What is the radius of the electron’s circular path?

Answer 1

energy in Joules = charge in Coulombs * Potential difference in Volts
= 1.6 * 10^-19 * 35 = 56 * 10^-19 Joules
so
(1/2) m v^2 = 56 * 10^-19
(1/2)(9.1*10^-31) v^2 = 56 * 10^-19
v^2 = 12.3 * 10^12
v = 3.5 * 10^6 m/s (luckily well below speed of light)
F = q V cross B = 1.6*10^-19 *3.5*10^6 * 0.75
V is East . B is down so V cross B is North
m v^2 / R = F

Answer 2

Whoops. q is NEGATIVE so q V cross B is SOUTH

Answer 3

Okay, makes more sense now. Thanks!