How Does Electric Field Influence Electron Movement in a Capacitor?

Click For Summary
SUMMARY

The discussion focuses on the influence of electric fields on electron movement within a capacitor. Key equations include the relationship between potential energy (PE) and kinetic energy (KE), specifically the conversion of PE gained by an electron to KE as it moves between capacitor plates. The electric field (E) is defined as E = ρ/ε, where ρ is charge density and ε is permittivity. The difference in potential energy between two points in the capacitor is calculated using U = qρd/ε, equating this with the kinetic energy at the initial point provides a method to determine electron velocity.

PREREQUISITES
  • Understanding of electric fields and their properties
  • Familiarity with kinetic and potential energy concepts
  • Knowledge of calculus for integrating forces
  • Basic principles of capacitor operation
NEXT STEPS
  • Study the derivation of electric field equations in capacitors
  • Learn about the relationship between charge density and electric field strength
  • Explore the concepts of work and energy in electric fields
  • Investigate projectile motion principles as they relate to electron movement
USEFUL FOR

Students studying physics, particularly those focusing on electromagnetism, electrical engineers, and anyone interested in the dynamics of charged particles in electric fields.

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: 409
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:

<br /> U = -W_{ext} = -\int F_{ext}\cdot dr<br />

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:

<br /> E = \frac{\rho}{\epsilon}<br />

and hence the force on the particle is given by:

<br /> F = \frac{q\rho}{\epsilon}<br />

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

<br /> U = \frac{q\rho d}{\epsilon}<br />

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
6K