Two positive charges of magnitude `q` are placed at the ends of a side ( side 1) of a square of side `2a` . Two negative charges of the same magnitude are kept at the other corners . Staring from rest , a charge `Q` moves from the middle of side `1` to the centre of square , its kinetic energy at the centre of square is -.

Options

A. `1/(4pi in_(0))(2qQ)/a(1-1/sqrt(15))`
B. zero
C. `1/(4pi in_(0))(2qQ)/a(1+1/sqrt(5))`
D. `1/(4pi in_(0))(2qQ)/a(1-1/sqrt(5))`

Answer

A

Explanation

`DeltaK=W=Q(V_(A)-V_(B))`
`=Q[((2kq)/a-(2kq)/(asqrt(5)))-0]`
`=(2KqQ)/(a)(1-1/sqrt(5))`