Question
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 $$\frac{Q}{q}$$ equals:
A.
$$-1$$
B.
$$1$$
C.
$$ - \frac{1}{{\sqrt 2 }}$$
D.
$$ - 2\sqrt 2 $$
Answer :
$$ - 2\sqrt 2 $$
Solution :
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 $$\sqrt 2 F' = - F$$
$$\eqalign{ & \sqrt 2 \times k\frac{{Qq}}{{{\ell ^2}}} = - k\frac{{{Q^2}}}{{{{\left( {\sqrt 2 \ell } \right)}^2}}} \cr & \Rightarrow \frac{Q}{q} = - 2\sqrt 2 \cr} $$
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 $$\sqrt 2 F' = - F$$
$$\eqalign{ & \sqrt 2 \times k\frac{{Qq}}{{{\ell ^2}}} = - k\frac{{{Q^2}}}{{{{\left( {\sqrt 2 \ell } \right)}^2}}} \cr & \Rightarrow \frac{Q}{q} = - 2\sqrt 2 \cr} $$
