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 figure. The point $B$ is at the lage 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 $A$. The minimum distance (in meter) of $+q$ from $+Q$ during the motion is $d(1+\sqrt{A})$. Find the value of $A$.
(Take $Qq=4\pi {\epsilon }_{0}m{u}^{2}d$ and $d=(\sqrt{2}-1)$ meter)

• A. $3$
• B. $2$
• C. $4$
• D. $5$

Answer 1

The correct option is B $2$

The path of the particle will be as shown in figure. At the point of minimum distance $\left(D\right)$ the velocity of the particle will be perpendicular to its position vector w.r.t $+Q$
$\frac{1}{2}m{u}^2+0=\frac{1}{2}m{v}^2+\frac{kQq}{r_{min}}.......\left(1\right)$
Torque on $q$ about $Q$ is zero hence angular momentum about $Q$ will be conserved
$\Rightarrow mv{r}_{min}=mud.......\left(2\right)$
By eqn $\left(2\right)$ in $\left(1\right)$
$\frac{1}{2}m{u}^2=\frac{1}{2}m{\left(\frac{ud}{r_{min}}\right)}^2+\frac{kQq}{r_{min}}$
$\Rightarrow \frac{1}{2}m{u}^2(1-\frac{d^2}{{r_{min}}^2})=\frac{m{u}^{2}d}{r_{min}}(\because kQq=m{u}^{2}d)$
$\Rightarrow {r_{min}}^2-2r_{min}d-d^2=0$
$\Rightarrow r_{min}=\frac{2d\pm \sqrt{4d^2+4d^2}}{2}=d(1\pm \sqrt{2})$
Distance cannot be negative
$\therefore r_{min}=d(1+\sqrt{2})$