An electron is released from rest in a uniform electric field. the electron accelerates vertically upward, traveling 4.50 m in the first [math]3.00 \mu s[/math] after it is released. (a) what are the magnitude and direction of the electric field? (b) are we justified in ignoring the effects of gravity? justify your answer quantitatively.

(a) 5.69 N/C, vertically downward

We can calculate the acceleration of the electron by using the SUVAT equation:

where

d = 4.50 m is the distance travelled by the electron

u = 0 is the initial velocity of the electron

is the time of travelling

a is the acceleration

Solving for a,

Given the mass of the electron,

We can find the electric force acting on the electron:

And the electric force can be written as

where

is the charge of the electron

E is the magnitude of the electric field

Solving for E,

The negative sign means that the direction of the electric field is opposite to the direction of the force (because the charge is negative): since the force has same direction of the acceleration (vertically upward), the electric field must point vertically downward.

(b) Yes

We can answer the question by calculating the magnitude of the gravitational force acting on the electron, to check if it is relevant or not. The gravitational force on the electron is:

where

is the mass of the electron

is the acceleration due to gravity

Substituting,

We see that the gravitational force is basically negligible compared to the electric force calculated in part (a), therefore we can say it is justified to ignore the effect of gravity in the problem.