Solving Electron Deflection in a Parallel Plate Apparatus

Click For Summary
SUMMARY

The discussion focuses on calculating the vertical deflection and exit velocity of an electron in a parallel plate apparatus with a potential difference of 600V. The setup measures 10 cm in length and 2.0 m in width, with the electron traveling horizontally at a speed of 8.0 x 10^7 m/s. Key equations used include the electric field calculation and the relationship between force, mass, and acceleration, specifically F=ma. The analysis confirms that gravity can be neglected due to its minimal effect compared to the electric force acting on the electron.

PREREQUISITES
  • Understanding of electric fields and forces
  • Familiarity with basic kinematics and projectile motion
  • Knowledge of the relationship between potential difference and electric field
  • Ability to apply Newton's second law (F=ma)
NEXT STEPS
  • Calculate the electric field produced by a 600V potential difference across 2.0 m
  • Determine the time taken for the electron to traverse the 10 cm length of the apparatus
  • Compute the vertical deflection of the electron using kinematic equations
  • Analyze the exit velocity of the electron in both magnitude and direction
USEFUL FOR

Physics students, educators, and anyone interested in understanding the dynamics of charged particles in electric fields, particularly in experimental setups involving parallel plate apparatuses.

lovemake1
Messages
147
Reaction score
1

Homework Statement



Question Reads:

An electron enters a parallel plate apparatus 10 cm long and 2.0 wide, moving horizontally at 8.0 10 ^7 m/s. if the potential difference across the plates is 600V, determine

a) the vertical deflection of the electron from its original path
b) the velocity with whih it leaves the parallel plate appratus, in both maginitude and direction

Homework Equations

,

v = sqrt ( 2qv / m )



The Attempt at a Solution



i have no clue where to start, but i guess this is like a projectile motion ? with gravity applied?
 
Physics news on Phys.org
lovemake1 said:

The Attempt at a Solution



i have no clue where to start, but i guess this is like a projectile motion ? with gravity applied?

Yes it is like projectile motion, except not with gravity.

If the pd is 600 V and the plates are 2m wide, what is the electric field produced?

Also if the electric field is the force exerted per unit charge and F=ma, what is the acceleration?
 
you can straight away neglect gravity. calculate & see yourself how small mg is in comparison to qE.

the elec field is perpendicular to the electron's initial velocity. say electron was initially moving along x & the field is along y.

now the filed accelerates the electron along y.
first, find out the time taken for the electron to leave the setup.
that is just length / vx. (vx = initial velocity. does not change)

so electron experiences an acceleration along y for this much time. so u can find vy.
 

Similar threads

Replies
4
Views
2K
  • · Replies 6 ·
Replies
6
Views
3K
Replies
1
Views
3K
Replies
3
Views
9K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 3 ·
Replies
3
Views
7K
  • · Replies 6 ·
Replies
6
Views
14K
Replies
4
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K