Electrostatics: Finding possible value of charge

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Homework Help Overview

The problem involves two charged particles with equal but opposite charges, differing masses, and specific initial velocities. The scenario describes their motion and separation over time, seeking to determine the possible values of their charges.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to apply conservation principles but questions how to determine the direction of the final velocity vectors in the context of the problem.

Discussion Status

The discussion includes requests for help, with some participants emphasizing the importance of patience and adherence to forum rules. There is no explicit consensus on the approach to the problem yet.

Contextual Notes

Participants are reminded to follow forum guidelines regarding posting frequency and attempts shown, indicating a structured approach to the discussion is expected.

utkarshakash
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Homework Statement


Two small particles have electric charges of equal magnitude and opposite signs. The masses of the particles are m and 2m. Initially, the distance between the particles is d, and the velocities of the particles have equal magnitude v. However, the velocity of particle 2m is directed away from particle m, whereas the velocity of particle m is directed perpendicular to the line connecting the particles. In the subsequent motion of the particles, they are found to be at a distance 3d from each other—twice! Find the possible values of the charge of each particle.


The Attempt at a Solution



I can apply conservation of energy and conservation of angular momentum for this problem. But how do I know what will be the direction of final velocity vectors?
 
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I need some help.
 
utkarshakash said:
I need some help.

Patience, patience.

It's too soon to bump your post.

Please, review the rules.
 
SammyS said:
Patience, patience.

It's too soon to bump your post.

Please, review the rules.

Are you saying that I have not shown any attempt towards the problem?
 
No what he is saying is that you posted into your own thread too soon.
 

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