Question

what is the magnitude and the direction of an electric field when an electron is accelrated vertically upward at a distance of 5.5 metres in 3 micro seconds?

Answer 1

First, E direction is the direction of a proton goes, so in this case, it is opposite of the electron path, so DOWNWARD is E.

distance=1/2 a t^2=1/2 Eq t^2
solve for E