How Does Electric Field Influence Electron Movement in a Capacitor?

Click For Summary

Homework Help Overview

The discussion revolves around understanding how an electric field influences electron movement within a capacitor. Participants are exploring the relationship between potential energy and kinetic energy as electrons move between the plates of the capacitor.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to calculate the potential energy gained by an electron and its corresponding kinetic energy, expressing uncertainty about their calculations. They also question how to determine the velocity of electrons emerging from one of the capacitor plates.
  • Some participants suggest relating the problem to projectile motion and clarify the role of initial velocity and forces acting on the electron.
  • Others discuss the potential energy differences between points in the capacitor and provide formulas for calculating work done by the electric field.

Discussion Status

The discussion is active, with participants providing insights and clarifications on the relationships between potential energy, kinetic energy, and the forces acting on the electron. While some guidance has been offered, there is no explicit consensus on the correct approach to the problem yet.

Contextual Notes

Participants are navigating the complexities of the problem, including the need for specific initial conditions such as initial velocity, which are not provided. There is an ongoing exploration of the assumptions related to the forces and energy transformations involved in the electron's movement.

Gunman
Messages
25
Reaction score
0

Homework Statement


I have attached the question.

Homework Equations


a)Potential Energy the electron gains = Kinetic energy it looses
1eV = 1.6 * 10^-19
Total P.E gained by the electron = 100 * 1.6 * 10^-19 = 1.6*10^-17
Therefore the kinetic energy it must possesses to reach the plate RS is 1.6 *10^-17 J.
I am wrong but I am not sure where I am incorrect. Any help would be appreciated. =)

b) I am not very sure on how to go about doing this part of the question as I would need the initial velocity, or the acceleration or time taken for it to reach B to find the velocity of electrons that emerge from B?

Thank you.
 

Attachments

  • Physics Qn.jpg
    Physics Qn.jpg
    21.8 KB · Views: 421
Physics news on Phys.org
You know the initial velocity is your variable, you know the constant force in the y direction, and no force in the x direction. It is akin to a projectile motion problem...
 
Yeah, got that part of the question. Thanks. =) But how do I go about doing the first part of the question?
 
The difference between Potential Energy at 'A' and that just beneath the 2nd plate [let's call this point P] is what will give you the Kinetic Energy at 'A'. This is because since the electron has no velocity at P and hence, it's KE is zero and the KE at 'A' is converted to the PE at 'P'. To calculate the PE difference, you can use the formula:

[tex] U = -W_{ext} = -\int F_{ext}\cdot dr[/tex]

Also, since the force acts in only one direction, i.e. the direction perpendicular to the plate, all points which are at the same height from the 1st plate will have the same PE. This is because if the electron were moved along a line of similar height, then no work will be done as F and r are perpendicular in this case, and it's dot product will be zero.

Inside a capacitor, at any point, the Electric field is given as:

[tex] E = \frac{\rho}{\epsilon}[/tex]

and hence the force on the particle is given by:

[tex] F = \frac{q\rho}{\epsilon}[/tex]

Since, the height difference of the two plates is 'd', we have the difference in PE as:

[tex] U = \frac{q\rho d}{\epsilon}[/tex]

Equate it with the KE at 'A', and you shall have your answer.
 

Similar threads

  • · Replies 5 ·
Replies
5
Views
1K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 12 ·
Replies
12
Views
2K
  • · Replies 3 ·
Replies
3
Views
9K
  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 9 ·
Replies
9
Views
8K
Replies
1
Views
3K
  • · Replies 7 ·
Replies
7
Views
7K