A positive charge `+Q` is fixed at a poibt A. Another positively charged particle of mass m and charge +q is projected from a point B with velocity u as shown in (Fig. 3.103). Point B is at a large distance from A and at distance d from the line A C. The initial velocity is parallel to the line A C. The point C is at a very large distance from A. Find the minimum distance (in meter) of +q from +Q during the motion. Take `Q q = 4 pi epsilon_0 m u^2 d` and `d (sqrt(2) - 1) m`.
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Explanation

The path of the particle will be as shown in (fig. 3.104). At the point of minimum distance D, the velocity the particle will be `_|_` to its position vector with respect to +Q. Applying conservation of angular momentum, we get
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`m u d = mvr_(min) or vr_(min) = u d`
Now by conservation of energy
`(1)/(2) m u^2 + 0 = (1)/(2) m v^2 + (K Q q)/(r_(min))`
or `(1)/(2) m u^2 (1 - (d^2)/(r_(min)^2) = (m u^2 d)/(r_(min))`
`[because KQ q = m u^2 d (given)]`
or `r_(min)^2 - 2r_(min) d - d^2 = 0`
or `r_(min) = (2 d +-sqrt(4 d^2 + 4 d^2))/(2) = d (1 +- sqrt(2))`
Since distance cannot be negative
`r_(min) = d (1 + sqrt(2))`.