Two identical particles of mass m and charge | vedclass

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Question

(b)

At distance of closest approach, Total initial $KE =$ Total final PE

$	herefore \quad \frac{1}{2} m v^2+\frac{1}{2} m v^2=\frac{k q^2}{r}$

Vedclass · Accepted Answer

$\frac{q^2}{4 \pi \varepsilon_0 m v^2}$

Vedclass · Answer

(b)

At distance of closest approach, Total initial $KE =$ Total final PE

$	herefore \quad \frac{1}{2} m v^2+\frac{1}{2} m v^2=\frac{k q^2}{r}$

$\Rightarrow \quad m v^2=\frac{q^2}{4 \pi \varepsilon_0 r}$

$\Rightarrow \quad r=\frac{q^2}{4 \pi \varepsilon_0 m v^2}$

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

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