Question

Two identical particles of mass m carry a charge Q each. Initially one is at rest on a smooth horizontal plane and the other is projected along the plane directly towards first particle from a large distance with speed $\upsilon$. The closest distance of approach will be

• A. $\frac{1}{4\pi {ϵ}_0}\frac{Q^2}{mv}$
• B. $\frac{1}{4\pi {ϵ}_0},\frac{4Q^2}{m{v}^2}$
• C. $\frac{1}{4\pi {ϵ}_0},\frac{2Q^2}{m{v}^2}$
• D. $\frac{1}{4\pi {ϵ}_0},\frac{3Q^2}{m{v}^2}$

Answer 1

The correct option is B $\frac{1}{4\pi {ϵ}_0},\frac{4Q^2}{m{v}^2}$


At the closest distance of approach the two particles must have the same velocities applying COE
$k_i+v_i=k_f+v_f\frac{1}{2}m{v}^2+\frac{k{Q}^2}{\infty }=\frac{1}{2}m{v}_1^2+\frac{1}{2}m{v}_1^2+\frac{k{Q}^2}{x}\Rightarrow \frac{k{Q}^2}{x}=\frac{1}{2}m{v}^2-m{v}_1^2\cdots \left(1\right)$
From COM,
$mv=m{v}_1+m{v}_1$
$\Rightarrow v_1=\frac{v}{2}\cdots \left(2\right)$
(2) in (1) gives
$\frac{k{Q}^2}{x}=\frac{m{v}^2}{4}\Rightarrow x=\frac{1}{4\pi {ϵ}_0},\frac{4Q^2}{m{v}^2}$