# Four charges are arranged at the corners of square ABCD. -2q at A, +2q at D, -q at C and +q at B....

Four charges are arranged at the corners of square ABCD.

-2q at A, +2q at D, -q at C and +q at B.

What would be the force on a charge +q which is kept at the centre of the square?

Options are-

A. Zero

B. Along diagonal AC

C. Along diagonal BD

D. Perpendicular to the side AB

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Charges are placed as (A,-2q),(B,q) ,(C,-q) and (D,2q),the vertices of

square ABCD. Let length of the side of square be r.

The force act B along side AB say F_A=k((-2q)(q))/r^2 (i)

The force act B along side CB say F_C=k((-q)(q))/r^2 (ii)

The resultant of force F_a and F_c say F_AC,

F_AC=sqrt((F_A)^2+(F_c)^2)) alog direction DB .

The force act B along BD say F_D=k((2q)(q))/r^2

F_AC=(k(q)^2/r^2)sqrt(5)

If resultant is zero then F_AC=F_D

But F_AC is not equal to F_D,therefore resultant is not equal to zero.

So unbalance force act along the direction of diagonal BD.

**So option C is corrcet answer.**