(a) 5.69 N/C, vertically downward
We can calculate the acceleration of the electron by using the SUVAT equation:
[tex]d=ut+\frac{1}{2}at^2[/tex]
where
d = 4.50 m is the distance travelled by the electron
u = 0 is the initial velocity of the electron
[tex]t=3.00 \mu s = 3.0 \cdot 10^{-6} s[/tex] is the time of travelling
a is the acceleration
Solving for a,
[tex]a=\frac{2d}{t^2}=\frac{2(4.50)}{(3.0\cdot 10^{-6})^2}=1.0\cdot 10^{12} m/s^2[/tex]
Given the mass of the electron,
[tex]m=9.11\cdot 10^{-31} kg[/tex]
We can find the electric force acting on the electron:
[tex]F=ma=(9.11\cdot 10^{-31})(1.0\cdot 10^{12})=9.11\cdot 10^{-19}N[/tex]
And the electric force can be written as
[tex]F=qE[/tex]
where
[tex]q=-1.6\cdot 10^{-19}C[/tex] is the charge of the electron
E is the magnitude of the electric field
Solving for E,
[tex]E=\frac{F}{q}=\frac{9.11\cdot 10^{-19}}{-1.6\cdot 10^{-19}}=-5.69 N/C[/tex]
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:
[tex]F=mg[/tex]
where
[tex]m=9.11\cdot 10^{-31} kg[/tex] is the mass of the electron
[tex]g=9.81 m/s^2[/tex] is the acceleration due to gravity
Substituting,
[tex]F=(9.11\cdot 10^{-31})(9.81)=8.93\cdot 10^{-30}N[/tex]
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.
