Question

Two identical particle $s$ of mass $m$ and charge $q$ are shot at each other from a very great distance with an initial speed $\upsilon$ The distance of closest approach of the se charges is

• A. $\frac{q^{2}}{8\pi\varepsilon_{o} m\upsilon^{2}}$
• B. $\frac{q^{2}}{4\pi\varepsilon_{o} m\upsilon^{2}}$
• C. $\frac{q^{2}}{2\pi\varepsilon_{o} m\upsilon^{2}}$
• D. $0$

Answer 1

Answer: (B)

At distance of closest approach, Total initial $KE =$ Total final $PE$
$\therefore \frac{1}{2}m\upsilon^{2} +\frac{1}{2}m\upsilon^{2} =\frac{kq^{2}}{r}$

$\Rightarrow m\upsilon^{2}=\frac{q^{2}}{4\pi\varepsilon_{o}r} $
$\Rightarrow r=\frac{q^{2}}{4 \pi \varepsilon_{0} m v^{2}}$