Movement of Charged Particles in a field

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Homework Help Overview

The discussion revolves around the movement of charged particles, specifically two electrons, and their speed as they move apart in an electric field. The problem involves calculating their speed when they are separated by a distance of 10^-8m after being released from a distance of 10^-9m.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss calculating the electric potential energy at two different distances and how the change in energy relates to kinetic energy. There are attempts to apply energy conservation principles to find the speed of the electrons.

Discussion Status

Some participants have provided calculations and results, suggesting that the approach to the problem is valid. However, there is an acknowledgment of potential calculation errors, and participants are encouraged to double-check their work. No explicit consensus has been reached on the correct answer.

Contextual Notes

Participants are operating under the assumption that the only forces acting on the electrons are due to their electric charges, and they are discussing the implications of their calculations without resolving the discrepancies in their results.

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Homework Statement


Two electrons are 10-9m Apart when they are released.What is their speed when they are 10-8m apart?



Homework Equations


Ek=-EE
EE=Kq1q2
-------
r


The Attempt at a Solution


Ive tried 1/2mv^=-(Ke(squared) - Ke(squared)
------------ -----------
r2 r1
And i don't seem to get the answer
By the way the answer to the question is 4.78x10^5 m/s
thanks for the help
 
Physics news on Phys.org
Calculate electric potential energy when they are 10^-9m apart, and then when they are 10^-8m apart. The change in energy has become kinetic energy, and you may calculate it from this.
 
I did that.. And i still get the wrong answer but its pretty close.. i guess its a calculation error.. i will go back and recheck thanks for the help...
 
Yeah doublecheck your calculation. I did it and I got something like 4.79e5, so I'm pretty sure that's how you do it.
 

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