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?
B. Along diagonal AC
C. Along diagonal BD
D. Perpendicular to the side AB
1 Answer | Add Yours
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.
1 Reply | Hide Replies ▲
thank you sir for your kind reply.
But the answer given is D
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