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 $Qq=4\pi {\in }_{0}m{u}^{2}d$ and $d=(\sqrt{2}-1)$ meter.

Answer 1

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 $\perp$ to its position vector with respect to +Q.
Now by conservation of energy:-
$\frac{1}{2}m{u}^2+0=\frac{1}{2}m{v}^2+\frac{KQq}{r_{min}}....\left(1\right)$
Torque on q about Q is zero, hence angular momentum about Q will be conserved.
$\Rightarrow mv{r}_{min}=mud.....\left(2\right)$
by putting (2) in (1) $\Rightarrow \frac{1}{2}m{u}^2=\frac{1}{2}m{\left(\frac{ud}{r_{min}}\right)}^2+\frac{KQq}{r_{min}}$
$\Rightarrow \frac{1}{2}m{u}^2(1-\frac{d^2}{r_{min}^2})=\frac{m{u}^{2}d}{r_{min}}$
$\{\because KQq=m{u}^{2}d(\text{given}\left)\right\}$
$r_{min}^2-2r_{min}\times d-d^2=0$
$\Rightarrow r_{min}=\frac{2d\pm \sqrt{4d^2+4d^2}}{2}=d(1\pm \sqrt{2})$ the distance cannot be negative
$\therefore r_{min}=d(1+\sqrt{2})$
$r_{min}=(\sqrt{2}-1)(1+\sqrt{2})\{\because \text{Given}d=(\sqrt{2}-1\left)\right\}$
$=2-1$
$=1m$