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the distance of closest approach [closed]

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

I tried it by conservation of energy;
$1/2 mv^2 = kQ^2 / r $
where $r $ is closest approach.

but given answer is twice of my answer i.e. given is $k4 Q^2 /mv^2$

Then I thought that there is something that when the other charge moves there is force already on the first charge as plane is smooth so it also starts to move and when relative velocities is 0 then it would be closest approach, but I am not able to find that cause force is variable nor I am able to apply center of mass concept .

asked Apr 3, 2017 in Physics Problems by physicsapproval (2,320 points)
closed Apr 4, 2017 by physicsapproval

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Best answer

This question has already been answered in Closest distance of approach when charged particle projected.

answered Apr 3, 2017 by sammy gerbil (27,948 points)
selected Apr 4, 2017 by physicsapproval
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