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JEE Main 2004 (AIEEE) · Electrostatic Potential Energy

Electrostatic Potential Energy question from JEE Main 2004 (AIEEE)

Q62JEE Main 2004 · AIEEEMedium

Electrostatic Potential Energy

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 of 2v2v the closest distance of approach would be

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Answer: (d)

  1. At closest approach: 12mv2=kQqr\frac{1}{2}mv^2 = \frac{kQq}{r}, so r∝1v2r \propto \frac{1}{v^2}.
  2. Speed 2v⇒r′=r42v \Rightarrow r' = \frac{r}{4}
  3. So the answer is (d) r/4r/4.
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