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

Answer

A

Explanation

Refer to Fig, below.
`AC = AD= a`
`AE = AF = sqrt(a^(2) + (2a)^(2)) = sqrt(5) a`
`BC = BD = BF = sqrt(2) a`

Potential at point `A` is
`V_(A) = 2 ((1)/(4pi in_(0)) (q)/(a)) - 2((1)/(4pi in_(0)) (q)/(a sqrt(5)))`
`= (1)/(4pi in_(0)) (2q)/(a) (1- (1)/(sqrt(5)))`
Potential at point `B` is
`V_(B) = 2 ((1)/(4pi in_(0)) (q)/(sqrt(2)a)) -2 ((1)/(4pi in_(0)) (q)/(sqrt(2) a)) = 0`.
`:. V_(A) - V_(B) = (1)/(4pi in_(0)) (2q)/(a) (1-(1)/(sqrt(5)))`