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Question

Two identical particles of mass m carry charge Q each. Initially one is at rest on a smooth horizontal plane and the other is projected along the plane directly towards first particle from a large distance with speed v. The closest distance of approach will be

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Solution

The correct option is **C** 14πε04Q2mv2

Due to repulsive force the other particle will start moving away. The velocity of the first particle will decrease while that of the other will increase. At the point of minimum distance between the two both the particles will be moving at same velocity. Let this velocity be u

So using conservation of momentum we get

mv=2mu or u=v2

The initial energy of the system is given as 12mv2

and the energy at the minimum distance is given as

12m(v2)2+12m(v2)2+14πϵoQ2R

Equating the two energies we get

12mv2=14mv2+14πϵoQ2R

or

14mv2=14πϵoQ2R

or

R=4Q24πϵomv2

Due to repulsive force the other particle will start moving away. The velocity of the first particle will decrease while that of the other will increase. At the point of minimum distance between the two both the particles will be moving at same velocity. Let this velocity be u

So using conservation of momentum we get

mv=2mu or u=v2

The initial energy of the system is given as 12mv2

and the energy at the minimum distance is given as

12m(v2)2+12m(v2)2+14πϵoQ2R

Equating the two energies we get

12mv2=14mv2+14πϵoQ2R

or

14mv2=14πϵoQ2R

or

R=4Q24πϵomv2

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