A charged particle q is shot towards another charged particle Q which is fixed. With a speed v , it approaches Q up to a closest distance r and then r - Physics MCQ AIEEE 2004 2026 | ExamTest.live

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A charged particle q is shot towards another charged particle Q which is fixed. With a speed v, it approaches Q up to a closest distance r and then returns. If q were given a speed 2 v, the closest distance of approach would be:

Solution & Step-by-step Explanation

At the distance of closest approach, all kinetic energy is converted to electrostatic potential energy. \frac{1}{2} m v^{2} = \frac{k Qq}{r} ⟹ r \propto \frac{1}{v ^{2}} If the speed is doubled (v \to 2 v), the new distance r^{'} will be: r^{'} = \frac{r}{2 ^{2}} = \frac{r}{4}

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A charged particle q is shot towards another charged particle Q which is fixed. With a speed v, it approaches Q up to a closest distance r and then returns. If q were given a speed 2 v, the closest distance of approach would be:

r

2 r

r /2

r /4

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