Electrostatics and infinately charged plate question driving me nuts

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Homework Help Overview

The problem involves an electron released from rest near an infinite charged plane, where it accelerates towards the plane and collides with a specified speed. The objective is to determine the surface charge density of the plane.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the relationship between velocity, distance, and acceleration, with some attempting to apply kinematic equations. There is uncertainty about how to incorporate the velocity into the calculations and whether certain equations are applicable for the scenario.

Discussion Status

Participants are actively exploring different approaches to relate the given variables. Some have suggested using kinematic equations, while others are questioning the appropriateness of certain formulas. There is no explicit consensus, but the discussion is progressing with various interpretations being examined.

Contextual Notes

There is a mention of confusion regarding the application of equations for constant versus accelerated motion, indicating a need for clarification on the assumptions being made in the problem setup.

ursulan
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Homework Statement


An electron is released from rest 2.0 cm from an infinite charged plane. It accelerates toward the plane and collides with a speed of 1.0x10^7m/s. What is the surface charge density of the plane?The answer is 2.52×10−7 C/m^2 !


Homework Equations


v=d/t, d=0.5at^2, E=n/2*8.85x10^-12, F=ma=Eq


The Attempt at a Solution


I don't know what to do with the velocity. I tried relating it to energy but got stuck.
 
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ursulan said:
I don't know what to do with the velocity.
Use that and the distance to find the electron's acceleration.
 
using v=d/t t is 2E-9
using a=v/t a is 5E15
E=ma/q E=28469
E=n/(2 epsilon not).. n=5.03E-7

which is 2 times too big, but isn't E=n/(2 epsilon not) the correct equation, with the 2?
 
ursulan said:
using v=d/t t is 2E-9
That equation applies for constant velocity, not accelerated motion. But you can use it if you replace v (the final velocity) with the average velocity. What's the average velocity?

You can also make use of additional kinematic relationships.
 
ahh so I guess you mean Vf^2=V0^2 + 2ad because that works haha

Thanks!
 
That's the one. :wink:
 

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