An electron is projected horizontally into the uniform electric field directed vertically downward between two parallelplates. The plates are 2.00 cm apart and are of length 4.00 cm. The initial speed of the electron is vi = 8.00 × 106 m/s. As it enters the region between the plates, the electron is midway between the two plates; as it leaves, the electron just misses the upperplate. a)What is the magnitude ofv the electric field?

455N/C

Explanation:

from  the question, the following data can be derived

Distance between plates=2cm=0.02m

length of plates=4cm=0.04m

initial speed of electron=8*10^6m/s

Note: the speed giving is the speed associated with the horizontal motion since it moves to cover the 4cm distance

we solve the equation component by component.

For the horizontal component, the time it takes to cover distance of 0.04m can be calculated as

this same time is used to cover the vertical distance which is midway between the plate,

Hence vertical distance covered is 0.02/2=0.01m

The acceleration in the vertical component can be calculated as

since

F=qE

also F=ma

then

qE=ma

E=(ma)/q

m=mass of electron=9.1*10^-32kq

q=charge of electron=1.6*10^-19c

a=acceleration

if we substitute values  

answer: its c lololo

  distance on the y-axis and time on the x-axis.

explanation:

the instantaneous velocity = change in position/time taken

the slope of the graph should give instantaneous velocity.

  slope of a graph = change in value of y -axis / change in values of x -axis

  comparing both the equations

    change in value of y -axis = change in position

  change in values of x -axis = time taken

so position values should be on the y axis and time values should be on the x axis.

distance on the y-axis and time on the x-axis.