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

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(4) r/4. At the closest approach the particle is momentarily at rest, so all of the kinetic energy has become electrostatic potential energy. 1/2mv²=1/(4πε₀)(qQ)/r, hence r=(qQ)/(4πε₀)·2/(mv²), i.e. r∝1/(v²). Replacing v by 2v multiplies v² by 4. So the new closest distance is r/4.

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

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Why the other options are wrong