Question 538: Electrostatics & Capacitance

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 $$Q/q$$ equals:

Accepted Answer

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
$$\sqrt 2 F' =  - F$$
$$\sqrt 2  	imes k{{Qq} \over {{\ell ^2}}} =  - k{{{Q^2}} \over {{{\left( {\sqrt 2 \ell } ight)}^2}}}$$
$$ \Rightarrow {Q \over q} =  - 2\sqrt 2 $$

Answer

$$-1$$

Answer

$$1$$

Answer

$$ - {1 \over {\sqrt 2 }}$$

Answer

$$ - 2\sqrt 2 $$

Electrostatics & Capacitance

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:

Choose the correct answer: