PLEASE HELP An electron is pushed eastwards with a force of 1.64 x 10-21 N when placed in an electric
field. Determine the
(a) magnitude and
(b) direction of the electric field.

1.02 x 10^-2 N/C to the east

Explanation:

Hi,

To find the magnitude of the electric field in this problem, you simply have to use the formula F = qE

F = Force

q = charge of the particle

E = magnitude of the electric field

Here, we are told that a force is being acted on an electron and we aren't told any specific quantity of charge for the electron. Therefore, we can assume that the charge of the electron will be its conventional charge of 1.602 x 10^-19 C

We are also given the force which is 1.64 x 10^-21 N

We solve for E

E = F/q

We plug in

E = (1.64 x 10^-21 N)/(1.602 x 10^-19 C)

E = 1.02 x 10^-2 N/C

The direction will be the east because the angle didn't change.

54 degrees incident angle = reflected angle

angle is from normal turned to line of ray