Calculating Minimum Distance of Approach for Charged Particles

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The discussion focuses on calculating the minimum distance of approach for a positively charged particle +q projected towards a fixed charge +Q. The setup involves charge +Q at point A on line AC, with charge +q starting from point B at a perpendicular distance d from point C. Participants suggest using conservation of angular momentum and mechanical energy to solve the problem, particularly when the initial velocity is parallel to AC. The relevance of point C and distance d is debated, with some concluding that they matter only if the initial velocity aligns with AC. Ultimately, the conservation principles help derive the minimum distance of approach effectively.
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A positive charge +Q is fixed at a point A on line AC. Another positively charged particle of mass m and charge +q is projected from a point B with velocity u. The point C is at large distance from A and B is situated at distance d perpendicular from point C from AC

Find the minimum distance of approach of +q towards +Q during motion.




Take Qq = 4\pi \varepsilon _{0} and d=\sqrt{2}-1



I tried at an instance where the velocity component directed towards +Q becomes zero. But I can't do anything about the perpendicular component of velocity as how it will increase the distance of particle from line AC.
 
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Hi AGNuke! :smile:
AGNuke said:
A positive charge +Q is fixed at a point A on line AC. Another positively charged particle of mass m and charge +q is projected from a point B with velocity u. The point C is at large distance from A, and B is situated at distance d perpendicular from point C from AC

Find the minimum distance of approach of +q towards +Q during motion.

Is that the complete question? :confused:

C (and the distance d) seems to have no relevance unless the initial velocity u is parallel to AC.

If so, use conservation of angular momentum (because … ?) :wink:
 
tiny-tim said:
If so, use conservation of angular momentum (because … ?) :wink:

Yeah. I got it. I conserved the angular momentum as well as mechanical energy (to get the velocity at closest distance of approach out of business) and got my answers.

Thanks a lot. :smile:
 
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