Proton deflection in a capacitor

Click For Summary
SUMMARY

The discussion focuses on calculating the sideways deflection of a proton traveling through a parallel-plate capacitor with a surface charge density of 1*10^-6 C/m² and a gap width of 2.0 cm. The proton's initial speed is 1.0*10^6 m/s. The key equations involved include the kinematic equation for vertical displacement, d = 0.5at², and the need to determine the force causing the vertical acceleration due to the electric field within the capacitor. The final answer is expected to be expressed in millimeters of displacement.

PREREQUISITES
  • Understanding of electric fields and forces in capacitors
  • Familiarity with kinematic equations in two dimensions
  • Basic knowledge of charge density and its effects on particle motion
  • Ability to perform unit conversions, particularly from meters to millimeters
NEXT STEPS
  • Calculate the electric field strength in a parallel-plate capacitor
  • Learn how to derive the force acting on a charged particle in an electric field
  • Explore the application of kinematic equations in two-dimensional motion
  • Investigate the relationship between charge density and electric field strength
USEFUL FOR

Students studying physics, particularly those focusing on electromagnetism and kinematics, as well as educators seeking to enhance their understanding of particle motion in electric fields.

badBKO
Messages
2
Reaction score
0

Homework Statement


A proton traveling at a speed 1.0*106 m/s of enters the gap between the plates of a 2.0-cm-wide parallel-plate capacitor. The surface charge densities on the plates are 1*10^-6 C/m2.

How far has the proton been deflected sideways when it reaches the far edge of the capacitor? Assume the electric field is uniform inside the capacitor and zero outside.


Homework Equations


d=.5at2


The Attempt at a Solution


1*10-4
 
Physics news on Phys.org
Thread moved to Intro Physics.

Welcome to the PF, badBKO. Could you please show us more of your work, including units? What force creates the sideways acceleration? How do you calculate the magnitude of that force?
 
The force that moved it to the right is not needed. There is no acceleration in the x deminsion, on lay in the y. The ending answer should be in millimeters of displacement.


I tried using kinematics in 2 dimension x and y
x1, y1, t1, etc

and the equations

y1 = y0 + vy(t1-t0) + .5ay(t1-t0)2

used the same equation for the x variable also.
 

Similar threads

Replies
3
Views
5K
Replies
2
Views
5K
Replies
3
Views
9K
  • · Replies 6 ·
Replies
6
Views
14K
  • · Replies 4 ·
Replies
4
Views
7K
  • · Replies 3 ·
Replies
3
Views
3K
Replies
11
Views
2K
Replies
16
Views
8K
  • · Replies 26 ·
Replies
26
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K