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Four charges are arranged at the corners of square ABCD. ...

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rocky.goyal | eNoter

Posted May 20, 2013 at 10:10 AM via web

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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

Tagged with electrostatics, science

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Top Answer

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pramodpandey | College Teacher | Valedictorian

Posted May 20, 2013 at 10:59 AM (Answer #1)

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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.

 

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rocky.goyal | eNoter

Posted May 21, 2013 at 1:32 PM (Reply #1)

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thank you sir for your kind reply.

But the answer given is D

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