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Ask iit jee aieee pet cbse icse state board community Discussion Response Post to: Electric Field
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NugoRama (4800)

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Olaaa!! Perrrfect answer. 802  [1197 rates]

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 It is clearly mentioned in the question that charge C is clamped ..

By observation we see that charge C should be between A and B.

Also note here that before introducing C ..A and B were repelling each othe..hence ..the charge C should attract A and B towards each other. ( finally A and B shud be at rest.)

Hence C shud be negatively charged.( A and B are positively charged a and 2q) ..let charge on C be -Q.

Now let C be at a distance x from A ..i.e. d-x distance from B.

C is clampled ..not bothering C ..

For A and B to be at rest ..the bet force on them should be zero.

Net force on A = kq(-Q)/x^2 + kq(2q)/d^2 = 0

                 hence ..     Q/x^2 = 2q/d^2 ......(1)

Net force on B = k(2q)(-Q)/ (d-x)^2 + k(2q)(q)/d^2 = 0

                 hence .. Q/(d-x)^2 = q/d^2 ..(2)

From (1) and (2)

                               Q/ (d-x)^2 = Q/ 2x^2

        2x^2 = (d-x)^2

        root2.x = d-x

x(1+root2) = d

x = d / ( 1+root2)

putting value of x in any of the above ..u'll get the value of Q.

 


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