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 dis

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 + + from +Q during the motion. Take $Qq = 4 π ϵ_{0} mu^{2} d$ and $d = (2 - 1)$ meter.

TestHub · Accepted Answer

At the point of minimum distance the velocity of the particle is $⊥$ to its position vector w.r.t. $+ Q$ .

By conservation of energy: $\frac{1}{2} m u^{2} + 0 = \frac{1}{2} m v^{2} + \frac{K Qq}{r _{m i n}} \dots (1)$

Torque on $q$ about $Q$ is zero, hence angular momentum about $Q$ is conserved:

$m v r_{m i n} = m u d \dots (2)$

by (2) in (1), with $K Qq = m u^{2} d$ :

$\frac{1}{2} m u^{2} = \frac{1}{2} m (\frac{u d}{r _{m i n}})^{2} + \frac{m u ^{2} d}{r _{m i n}}$

$\Rightarrow r_{m i n}^{2} - 2 r_{m i n} d - d^{2} = 0$

$\Rightarrow r_{m i n} = \frac{2 d \pm 4 d ^{2} + 4 d ^{2}}{2} = d (1 \pm 2)$

distance cannot be negative, so $r_{m i n} = d (1 + 2)$

$= (2 - 1) (1 + 2) = 1$ m

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