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

pramodpandey | Student

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 


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.