Electrostatics

JEE Main level, single correct, medium

232 of 311

Question 232 of 311

A charged particle qq is shot towards another charged particle QQ which is fixed, with a speed vv it approaches QQ upto a closest distance rr and then returns. If qq were given a speed 2v2 v, the closest distances of approach would be

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Answer: (D) r/4\mathrm{r} / 4

By principle of conservation of energy
Finally, 12m(2v)2=KqQr2\frac{1}{2} m(2 v)^2=\frac{K q Q}{r^2}
Equation (i) ÷\div (ii),
14=r′r⇒r′=r4.\begin{aligned} & \frac{1}{4}=\frac{r^{\prime}}{r} \\ & \Rightarrow r^{\prime}=\frac{r}{4} . \end{aligned}

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