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

• A. 3
• B. 2
• C. 4
• D. 5

Answer 1

Answer: (B)

The path of the particle will be as shown in figure. At the point of minimum distance (D) the velocity of the particle will be $\perp$ to its position vector w.r.t. $+Q$.
$\frac{1}{2}m{u}^2+0=\frac{1}{2}m{v}^2+\frac{kQq}{r_{\min }}\dots \dots \dots \dots (1)$

$\because$ Torque on $q$ about $Q$ is zero hence angular momentum about Q will be conserved.
$\Rightarrow mv{r}_{\text{min }}=$ mud ..............(2)
By eq. (2) in eq. (1)
$\frac{1}{2}m{u}^2=\frac{1}{2}m{\left(\frac{ud}{r_{\min }}\right)}^2+\frac{kQq}{r_{\min }}$
$\Rightarrow \frac{1}{2}m{u}^2\left(1-\frac{d^2}{r_{\min }^2}\right)=\frac{m{u}^{2}d}{r_{\min }}\left\{\because kQq=m{u}^{2}d\right\}$
$\Rightarrow r_{\min }^2-2r_{\min }d-d^2=0$
$\Rightarrow r_{\min }=\frac{2d\pm \sqrt{4d^2+4d^2}}{2}=d(1\pm \sqrt{2})$
$\because$ distance cannot be negative
$\therefore r_{\min }=d(1+\sqrt{2})$