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 .

Options

A. `1/(4pi epsilon_(0))Q^(2)/(mv)`
B. `1/(4piepsilon_(0))(4Q^(2))/(mv^(2))`
C. `1/(4piepsilon_(0))(2Q^(2))/(mv^(2))`
D. `1/(4piepsilon_(0))(3Q^(2))/(mv^(2))`

Answer

B

Explanation

By mechanical energy conservation
`(PE+KE)_(i)=(PE+KE)_(f)`
`0+1/2 mv^(2)+0=(KQ^(2))/d+1/2 m (v/2)^(2)xx2`
( `:'` from momentum conservation at closet approach, both particle will move with a common speed `v//2`)
`:. D=(4KQ^(2))/(mv^(2))`