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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