Question

an electron enters the region of a uniform,electric field with 3X10^6m/s and e=200 N/C . the horizontal length of the plates is 0.100m. a) find the acceleration of the electron while it is in the electric field . b) if the electron enters the field at time t=0 find the time at which it leaves the field c) the vertical position of the electron as it enters the field is 0 , what is its vertical position when it leaves the field?

Answer 1

charge of electron = 1.6 * 10^-19 Coulombs
so force on electron = 200 Newtons / Coulomb * 1.6 * 10^-19 Coulombs
= 3.2 * 10^-17 Newtons
Now it is a freshman Physics problem
F = m a
3.2 * 10^-17 Newtons = 9.11*10^-31 kg * a
a= (3.2/9.11) 10^14 m/s^2 = 3.51*10^13 m/s^2 (part a)
well if it goes 0.1 meters at 3*10^6 m/s it takes (0.1 / 3) *10^-6 seconds
= 3.33 * 10^-8 seconds (hard to time with stopwatch)
h = (1/2) a t^2 = 1.76*10^13 * 11.1 *10^-16 = 19.5 *10^-3 or about 20 millimeters or 2 centimeters (less than an inch)