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 . Starting from rest , if 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. zero
B. `1/(4piepsilon_(0)) (2qQ)/a (1+1/(sqrt(5)))`
C. `1/(4piepsilon_(0))(2qQ)/a ( 1 - 1/(sqrt(5)))`
D. `1/(4piepsilon_(0)) (2qQ)/a (1-1/(sqrt(5)))`

Answer

D

Explanation

Work done on charge Q = change in K.E = `E_(f) -E_(f) = E_(f) - 0 `
` :. Q (V_(A) - V_(B)) = E_(f)`
`Q (2/(4piepsilon_(0)) q/a -2/(4piepsilon_(0)) q/(sqrt(5)a) - 0 ) = E_(f)`
` :. E_(6) = (2qQ)/(4piepsilon_(0)a) (1-1/(sqrt(5)))`