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Four charges are arranged at the corners of square ABCD. ...
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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
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.
Posted by pramodpandey on May 20, 2013 at 10:59 AM (Answer #1)
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