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A charge particle 'q' is shot towards another charged particle 'Q' which is fixed , with a speed 'v'.It approaches 'Q' upto a closest distance r and then returns . If q were given a speed of '2v' the closest distance of approach would be

$(a)\;\large\frac{r}{2}\qquad(b)\;2r\qquad(c)\;r\qquad(d)\;\large\frac{r}{4}$

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Answer : (d) $\;\large\frac{r}{4}$
Explanation :
$\large\frac{1}{2}\;mv^2=\large\frac{kqQ}{r}---(1)$
$\large\frac{1}{2}\;m(2v)^2=\large\frac{kqQ}{r^{|}}$
$4 r^{|} = r $
$r^{|} = \large\frac{r}{4}\;.$
answered Feb 6, 2014 by yamini.v
 

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