How to calculate vertical electron deflection between two charged plates?

Click For Summary
To calculate vertical electron deflection between two charged plates, first determine the charge density using the equation E = sigma / (2 * epsilon_0). The vertical electric field causes a constant acceleration on the electron, similar to projectile motion problems on Earth. To find the vertical deflection, use kinematic equations that account for this constant acceleration over the horizontal distance traveled. The vertical component of the final velocity can also be calculated using these principles. Understanding the relationship between the electric field and motion will clarify the process of calculating vertical deflection.
Molly
Messages
3
Reaction score
0

Homework Statement


In the problem, an election is moving though 2 charged plates
vertical electric field amount is given and its perpendicular to the electron’s original horizontal velocity. The initial speed of the electron is given and the horizontal distance it travels is given. Then it asks for the charge density, the vertical deflection, and the vertical component of Vf.

Homework Equations


E=sigma/ 2* epsilon 0.

The Attempt at a Solution


So i think I got the charge density correct, hopefully. I used the above equation to calculate sigma. What I am stuck on is how do I calculate vertical deflection? I feel like this is something really stupid to be confused about, but I don't know how to start it.
 
Last edited by a moderator:
Physics news on Phys.org
Molly said:
how do I calculate vertical deflection?
It's a uniform field so produces constant acceleration, just like in Earth's surface trajectory problems.
Does that help?
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

Similar threads

Replies
11
Views
3K
Replies
1
Views
1K
  • · Replies 5 ·
Replies
5
Views
5K
  • · Replies 26 ·
Replies
26
Views
3K
  • · Replies 58 ·
2
Replies
58
Views
5K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
12
Views
2K
  • · Replies 9 ·
Replies
9
Views
2K