Answer to Question #100512 in Electricity and Magnetism for Harsha

Question

3-11. At some instant the velocity components of an electron moving between two charged parallel
plates are vx = 1.5105 m/s and vy = 0.30104 m/s. If the electric field between the plates is given by
E = j1.2104 N/C, (a) what is the acceleration of the electron? (b) When the x-coordinate of the
electron has changed by 2.0 cm what will be the velocity of the electron?

Answer 1

We are given

V_x= 1.5*10⁵ m/s V_y=0.3*10⁴ m/s

a) As electric field is in the positive j then the force acting on the electron will be in negative j direction

F=eE

ma=eE

a=eE/m

a=1.6*10^-19*1.2*10⁴/(9.1*10^-31)

a=2.1*10¹⁵ m/s² in negative j

b) as in positive x axis there is no electric field so no force will act on the electron

so time taken by the electron to travel 2 cm is 2*10^-3/(1.5*10⁵) = 1.8*10^-8sec

velocity in the x direction will remain to be the same

velocity in y= u_y-at

=0.3*10⁴ - 2.1*10¹⁵*1.8*10^-8

=0.3*10⁴ - 3.78*10⁷

=3.78*10⁷ in negative j