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Two identical particles of mass m carry a 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 be .

A

`(1)/( 4 pi epsi_(0)) (Q^(2))/(m v)`

B

`(1)/(4 pi espi_(0)) (4Q^(2))/(m v^(2))`

C

`(1)/(4 pi epsi_(0)) (2Q^(2))/(m v^(2))`

D

`(1)/(4 pi epsi_(0))(3Q^(2))/(m v^(2))`

Text Solution

Verified by Experts

The correct Answer is:
B
At closest distance they will have same speed u. By conservation of momentum,
`mv=mu+mu, u=(v//2)`
by cons. Of energy `k (QO)/d+1/2 mu^2+1/2 mu^2=1/2 mv^2, d=(4 k Q^2)/(m v^2)`
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