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 $A C$. The initial velocity is parallel to the line $A C$. 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 $Oq =4 \pi \varepsilon_0 mu ^2 d$ 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. $+Q$.
$\frac{1}{2} mu ^2+0=\frac{1}{2} mv ^2+\frac{ kQq }{ r _{\min }} \ldots \ldots \ldots \ldots(1)$

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