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?

 

[HalfLostAndSearching]

I appreciate your effort in solving this problem. Here are my responses to your questions:

1) In this case, the negative sign is appropriate since we are dealing with an electron. It is important to be consistent with the signs in our calculations in order to obtain accurate results.

2) The equation E = Q/Aε0 gives the magnitude of the charge, but it does not indicate which plate has which charge. In this case, we can determine the charge on the upper plate by considering the direction of the electric field. Since the electric field vector was found to be positive, the charge on the upper plate must also be positive.

3) Your work seems to be correct. Your solution for the electric field vector and potential difference are consistent with the given information. However, it would be helpful to also include the units in your calculations, as this can provide a better understanding of the results.

Overall, your approach and solution to this problem are correct. Keep up the good work!