JEE MAIN - Physics (2009 - No. 15)
A charge $$Q$$ is placed at each of the opposite corners of a square. A charge $$q$$ is placed at each of the other two corners. If the net electrical force on $$Q$$ is zero, then $$Q/q$$ equals:
$$-1$$
$$1$$
$$ - {1 \over {\sqrt 2 }}$$
$$ - 2\sqrt 2 $$
Explanation
Let $$F$$ be the force between $$Q$$ and $$Q.$$ The force between $$q$$ and $$Q$$ should be attractive for net force on $$Q$$ to be zero. Let $$F'$$ be the force between $$Q$$ and $$q.$$ For equilibrium
$$\sqrt 2 F' = - F$$
$$\sqrt 2 \times k{{Qq} \over {{\ell ^2}}} = - k{{{Q^2}} \over {{{\left( {\sqrt 2 \ell } \right)}^2}}}$$
$$ \Rightarrow {Q \over q} = - 2\sqrt 2 $$
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