Gujarati
Hindi
1. Electric Charges and Fields
normal

A charged particle $'q'$ is shot from a large distance with speed $v$ towards a fixed  charged particle $Q$. It apporaches $Q$ upto a closet distance $r$ and then returns. If $q$ were given a speed $'2v$', the closest distance of approach would be 

A

$r$

B

$2r$

C

$r/2$

D

$r/4$

Solution

$\frac{1}{2} m v^{2}=\frac{k Q q}{r} \Rightarrow \frac{1}{2} m(2 v)^{2}=\frac{k q Q}{r^{\prime}} \Rightarrow r^{\prime}=\frac{r}{4}$

Standard 12
Physics

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