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Three charges +Q, q, + Q are placed respectively, at distance, 0, d/2 and d from the origin, on the x-axis. If the net force experienced by + Q, placed at x= 0, is zero, then value of q is :

(1) +Q/2

(2) –Q/2

(3) –Q/4

(4) +Q/4

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Ans: (3) –Q/4

Sol: 

For equilibrium, 

$\vec{F}_a+\vec{F}_b=0$

$\vec{F}_a=-\vec{F}_b$

${kQQ\over d^2}=- {kQq\over (d / 2)^2}$

$\implies q=-{Q\over4}$

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