Proton deflection in a capacitor

AI Thread Summary
A proton traveling at 1.0*10^6 m/s enters a 2.0-cm-wide parallel-plate capacitor with surface charge densities of 1*10^-6 C/m². The discussion focuses on calculating the sideways deflection of the proton as it exits the capacitor, emphasizing the need to determine the force causing the vertical acceleration. Participants highlight the importance of showing work and using appropriate units in calculations. The correct approach involves applying kinematic equations to analyze motion in both the x and y dimensions. The final answer for deflection should be expressed in millimeters.
badBKO
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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
 
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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.
 
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