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Question

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=4π0mu2d and d=(21) meter.

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Solution


The path of the particle will be as shown in the figure. At the point of minimum distance (D), the velocity of the particle will be to its position vector with respect to +Q.
Now by conservation of energy:-
12mu2+0=12mv2+KQqrmin....(1)
Torque on q about Q is zero, hence angular momentum about Q will be conserved.
m vrmin=m ud.....(2)
by putting (2) in (1) 12mu2=12m(udrmin)2+KQqrmin
12mu2(1d2r2min)=mu2drmin
{KQq=mu2d(given)}
r2min2rmin×dd2=0
rmin=2d±4d2+4d22=d(1±2) the distance cannot be negative
rmin=d(1+2)
rmin=(21)(1+2){Given d=(21)}
=21
=1 m

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