A particle of mas s m carrying charge 'q' is projected with velocity ‘v’ from point 'A' towards an infinite line of charge from a distance 'a'. Its speed reduces to zero momentarily at point 'B' which is at a distance a/2 from the line of charge. If another particle with mass m and charge '-q' is projected with the same velocity 'v' from 'A' towards the line of charge. If v=102m/s, what will be its speed (in m/s) at 'B'?
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Detailed Solution
Applying conservation of mechanical energyΔK+ΔU=0 or Kf−Ki+qVf−Vi=0 ………(i)When a positive charge is projected towards the wire0−12mv2+qΔV=0⇒12mv2=qVB−VA …….(ii)When a negative charge is projected towards the wire 12mvB2−12mv2+(−q)ΔV=0⇒12mvB2−12mv2=qVB−VA ………(iii)From (ii) and (iii), we get 12mv2=12mvB2−12mv2⇒vB2=2v2 or vB=2v =2×102=20m/s
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A particle of mas s m carrying charge 'q' is projected with velocity ‘v’ from point 'A' towards an infinite line of charge from a distance 'a'. Its speed reduces to zero momentarily at point 'B' which is at a distance a/2 from the line of charge. If another particle with mass m and charge '-q' is projected with the same velocity 'v' from 'A' towards the line of charge. If v=102m/s, what will be its speed (in m/s) at 'B'?