Electron move through capacitor

[Matt Q.]

Homework Statement

An electron enters a region between two capacitor plates with equal and opposite charges. The plates are L=0.1 m by d=0.05 m, and the gap between the plates is h=0.002 m. During a short time interval of dt=0.002 s, while between the plates and far from the edges, the change of momentum is dp=<0,-9x10^-17,0> kg m/s. Ignore gravitational effects here since they are so weak compared to electric effects.
(a) What is the electric field vector between the plates (far from the edges)? (hint: momentum principle)
(b) What is the charge (magnitude and sign) on the upper plate?
(c) What is the potential difference dV = V_top - V_bottom ?

I figured out how to solve the problem, however, I have some questions.
1) Do I use 1.6x10^-19 or do I use -1.6x10^-19 ? I know the negative makes more sense since it's an electron but my friend were arguing that signs does not matter.
2) How do I tell if the charge I got is for the upper plate or lower plate? I used the equation: E = Q / A epsilon₀ to find the charge Q.
3) I'm not entirely sure that my work is correct. I might have done something wrong somewhere.

Homework Equations

a)
dp = F dt => F = dp / dt
F = qE => E = F / q

b)
E = Q / A epsilon₀ => Q = E A epsilon₀

c)
dV = -E dl

The Attempt at a Solution

a)
F = -9x10^-17 / 0.002 = -4.5x10^-14

E = 4.5x10^-14 / -1.6x10^-19 = 2.8x10⁵ V/m (not sure if this should be plus or minus because the sign of charge of electron)

b)
Q = 2.8x10⁵x0.1x0.05xepsilon₀ = 1.24x10^-8 ( not sure if this charge belongs to upper plate or lower plate )

c)
dV = -2.8x10⁵ x 0.002 = -562.5 Volt

 

[rude man]

Since no coordinate system was furnished with the problem it doesn't matter which signs you use. Furthermore, the sign of the E field, that of the upper plate charge, as well as that of the potential difference are all indeterminate.

 

[BvU]

Rudy's correct if you're really critical. I'm inclined to see the mentioning of a negative dp_y as an indication that you are supposed to consider the electron is pulled towards the bottom plate. I.e. take the positive y-direction upwards.

That way you get $\vec E$ pointing upwards and a negative V_top - V_bot.

The expression you use in b) is for Q on both plates, meaning Q_bot is + what you found and Q_top = - idem.

(physically, all you found is a charge difference, but because the exercise mentions equal and opposite charges, there is no doubt left).

My compliments for a clear and well-formulated post.

 

[Simon Bridge]

Off the specific questions:

(a) the literal calculaton would give a minus sign because you have a positive number divided by a negative number ... so keep the signs.
Think about the role the sign plays in the electric field. Remember, E is a vector - you are asked for the vector, so you answer should be a vector - vectors have a magnitude and a direction. What is the magnitude in this case? What is the direction?

(b) If the charge on the upper plate is Q then the charge on the lower plate is -Q ... does it matter to the problem description which is which?