A beam of electrons is shot into a uniform downward electric field of magnitude

Question

A beam of electrons is shot into a uniform downward electric field of magnitude 1.11 103 N/C. The electrons have an initial velocity of 1.05 107 m/s, directed horizontally. The field acts over a small region, 5.00 cm in the horizontal direction.
(a) Find the magnitude and direction of the electric force exerted on each electron.
I know the direction is upward, but I don't know how to calculate the magnitude.
(b) How does the gravitational force on an electron compare with the electric force?
I know the answer to B is The gravitational force is much smaller than the electric force.
(C)How far has each electron moved in the vertical direction by the time it has emerged from the field?
(d) What is the electron's vertical component of velocity as it emerges from the field? (Up is the positive y-direction.
(e) The electrons move an additional 20.6 cm after leaving the field. Find the total vertical distance that they have been deflected by the field.

Please when you answer these questions please tell me the formula and how you did each problem step by step. No matter what I do, I'm getting the answer wrong.

Explanation / Answer

(You should know that the equation editor did not work when I wrote this. Therefore the equations are very low-fi. Sorry.) (a)
From Coulombs law in our case
F = eE.................................(1), where e is the electron charge in Coulomb and E is the magnitude of the electrical field in Newton per Coulomb. Thus with e = 1.6 x 10^(-19) C we get F = 1.78 x 10^(-16) N.
If you are certain that the direction is upward I will not dispute that. However you should know the direction of the force due to an electric field is the same direction as the electric field, which can be seen in (1) as the force F an electric field E are the only vector quantities. That's why I think the direction is downward as the electric field is downward.
(b) Let calculate the gravitational force on the one electron
F_grav = m_electron x g = 9.11 x 10^(-31) x 9.80 m/s^2 = 8.93 x 10^(-30) N
From this and (a) we see that (F_elec) / (F_grav) = 1.99 x 10^(13) N,
Thus we conclude that the gravitational force is much smaller than the force due to the electric field. (c) I suppose that the gravitational force is of a magnitude so small that the problem suggests that it is to be neglected.
We first use Newtons second law to calculate the acceleration due to the electric field: a_elec = F_elec / m_electron = 1.95 x 10^(14) m/s^2. Now the velocity of the electrons is constant thus an electron will clear a distance of 5 cm in v = s/t => t = s / v = 4.76 x 10^(-9) s. Since the electric field is uniform/constant it implies the force due to the electric field it constant which implies the acceleration due to the electric field is constant hence the relation
s = 0.5at^2 + (v_init)t + s_init..................(2) applies. We set v_init = 0 m/s and s_init = 0 (v_init = zero as the is the initial velocity in the y direction and s_init = 0 means that I set the point of reference where the electron are before they enter the electric field.). Thus obtaining s = 0.22 cm,
which is how far one electron has moved vertically after clearing the field. I tried adding the acceleration due to gravity. Since a_elec is so much bigger than g it does not alter the result in a significant way.
(d) Here the answer depends on the answer in (a). You say it is upward and I believe it is downward.
In (c) we know that the electron clears the field in t = 4.7 x 10^(-9) s. And since the acceleration is constant the relation
v(t) = (a_elec)t + v_init applies. In (c) we set v_init = 0. Thus the vertical velocity component after an electron has cleared the field is
v_vertical = 92.82 x 10^4 m/s x r, where r is unit directional vector which in my opinion is -j. However given your answer to question (a) r is j = (0,1) (e) We just add the distance found in (c) to 20.6 cm. Thus s_total = 0.22 cm + 20.6 cm = 20.82 cm. Hope it helps.

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