# Closest distance of approach

1. Jul 11, 2012

### AGNuke

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.

2. Jul 11, 2012

### tiny-tim

Hi AGNuke!
Is that the complete question?

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 … ?)

3. Jul 11, 2012

### AGNuke

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.