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2. Electric Potential and Capacitance
medium
Two identical particles of mass $m$ and charge $q$ are shot at each other from a very great distance with an initial speed $v$. The distance of closest approach of these charges is
A
$\frac{q^2}{8 \pi \varepsilon_0 m v^2}$
B
$\frac{q^2}{4 \pi \varepsilon_0 m v^2}$
C
$\frac{q^2}{2 \pi \varepsilon_0 m v^2}$
D
$0$
(KVPY-2010)
Solution
(b)
At distance of closest approach, Total initial $KE =$ Total final PE
$\therefore \quad \frac{1}{2} m v^2+\frac{1}{2} m v^2=\frac{k q^2}{r}$
$\Rightarrow \quad m v^2=\frac{q^2}{4 \pi \varepsilon_0 r}$
$\Rightarrow \quad r=\frac{q^2}{4 \pi \varepsilon_0 m v^2}$
Standard 12
Physics
