Calculating Vertical Deflection of Electron in Parallel Plate Setup

[XJellieBX]

Question:
An electron with a velocity of 3.00x10^6m/s[horizontally] passes through two horizontal parallel plates, as in the attached diagram. The magnitude of the electric field between the plates is 120N/C. The plates are 4.0cm across. Edge effects are negligible.
a) Calculate the vertical deflection of the electron.

I'm not sure what I should do here and I would appreciate any help. I think it might be [tex]\epsilon[/tex]=[tex]\frac{kq}{r^2}[/tex], but I'm not exactly sure. Thanks.

 

[Doc Al]

What's the force on the electron while it's between the plates? The resulting acceleration? How long does it spend between the plates?

I think it might be [tex]\epsilon[/tex]=[tex]\frac{kq}{r^2}[/tex], but I'm not exactly sure.

That's an equation for the field from a point charge, which is not relevant here. You are given the field, so find the force.

 

[ogb p]

I can provide a response to the content and guide you through the process of calculating the vertical deflection of the electron in this parallel plate setup.

First, let's start by identifying the relevant variables and equations. The variables we have are the electron's velocity (3.00x10^6m/s), the electric field (120N/C), and the distance between the plates (4.0cm). The equation we will use is F=qE, where F is the force on the electron, q is the charge of the electron, and E is the electric field.

To calculate the vertical deflection, we need to find the force on the electron in the vertical direction. Since the electric field is in the horizontal direction, there will be no vertical force from the electric field. However, there will be a vertical force due to the electron's motion, which we can calculate using the equation F=ma, where m is the mass of the electron and a is the acceleration in the vertical direction.

To find the acceleration, we can use the equation a=v^2/r, where v is the velocity of the electron and r is the radius of the circular path it will follow due to the vertical force. In this case, the radius is equal to the distance between the plates (4.0cm).

Now, we can plug in all the values we have into the equations. The charge of an electron is 1.6x10^-19 C, and the mass is 9.11x10^-31 kg. So, the force on the electron in the vertical direction is:

F= (1.6x10^-19 C)(3.00x10^6 m/s)^2/4.0x10^-2 m
= 1.44x10^-13 N

Next, we can use this force to calculate the acceleration in the vertical direction:

a= 1.44x10^-13 N/9.11x10^-31 kg
= 1.58x10^17 m/s^2

Finally, we can use the equation for vertical deflection, d=1/2at^2, where d is the vertical deflection, a is the acceleration, and t is the time it takes for the electron to pass between the plates. Since the plates are 4.0cm across and the electron is traveling at 3.00x10^6 m/s, it