A positive charge +Q is fixed at a point A. Another positively charged particle of mass m and charge +q is projected from a point B with velocity u as shown in the figure. The point B is at large distance from A and at distance ‘d’ from the line AC. The initial velocity is parallel to the line AC. The point C is at very large distance from A. Find the minimum distance (in meter) of +q from +Q during the motion. Take Qq = 4πε 0 mu 2 d and 
meter.
Text Solution
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(1)
Sol. The path of the particle will be as shown in the figure. At the point of minimum distance the velocity of the particle will be ⊥ to its position vector w.r.t. +Q.
Now by conservation of energy:-
mu 2 + 0 = 
mv 2 + 
......
Torque on q about Q is zero, hence angular momentum about Q will be conserved
⇒ m v r min = m ud ......
∴ By putting in ⇒ 
mu 2 = 
m 
+ 
⇒ 
mu 2 
= 
{ KQq = mu 2 d }
⇒ r 2 min – 2r min d – d 2 = 0
⇒ r min = 
= d (1 ± 
)
Distance cannot be negative
∴ r min = d(1 + 
) = 1 m
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