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 epsilon_0)(2qQ)/a(1-1/(sqrt5))`
B. Zero
C. `1/(4pi epsilon_0)(2qQ)/a(1+1/(sqrt5))`
D. `1/(4pi epsilon_0)(2qQ)/a(1-1/(sqrt5))`

Answer

A

Explanation

Initial potential energy of the system `U_i =(2k qQ)/a+ (2k(-q)Q)/(sqrt5 a) =1/(4 pi e_0) (2q Q)/a [1-1/sqrt3]`
Final potential energy of the system, `U_f=0`
By conservation law of energy Gain in KE = Loss in PE
`K=1/(4 pi e_0) (2q Q)/a [1-1/sqrt5]`