An electron is released from rest in a uniform electric field

[hitek131]

I need help with this question if anyone can give me an idea of what to do. Any help would be greatly appreciated.

An electron is released from rest in a uniform electric field. The electron accelerates vertically upward, traveling 4.50 m in the first 3.00 micro seconds after it is released.
What is the magnitude of the electric field?

 

[Tide]

HINT: [itex]s = a t^2/2[/itex]

 

[quicknote]

I can provide you with a mathematical approach to determine the magnitude of the electric field in this scenario. Let's start with the equation for the acceleration of an object in a uniform electric field:

a = qE/m

In this equation, a represents the acceleration, q is the charge of the electron, E is the electric field strength, and m is the mass of the electron. We know that the electron is accelerating vertically upward, so we can use the equation for displacement to calculate the acceleration:

a = 2Δy/t^2

In this equation, Δy represents the displacement (4.50 m) and t represents the time (3.00 microseconds or 3.00 x 10^-6 seconds). Now we can set these two equations equal to each other and solve for the electric field strength:

qE/m = 2Δy/t^2

E = (2Δyq)/mt^2

Plugging in the known values for the charge of an electron (q = -1.60 x 10^-19 C) and the mass of an electron (m = 9.11 x 10^-31 kg), we get:

E = (2 x 4.50 m x -1.60 x 10^-19 C)/(9.11 x 10^-31 kg x (3.00 x 10^-6 s)^2)

E = -3.55 x 10^9 N/C

Therefore, the magnitude of the electric field in this scenario is 3.55 x 10^9 N/C. It is important to note that the negative sign indicates that the electric field is directed downward, which is consistent with the electron's acceleration being upward. I hope this helps you with your question. Let me know if you have any further inquiries.