A charge Q is placed at each of the opposite corners of a square. A… | JEE Main 2009 (AIEEE)

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Electric Charge & Coulomb's Law

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:

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Answer: (d)

Side $a$ . Force on $Q$ from the opposite $Q$ (distance $\sqrt{2}a$ ): $\dfrac{kQ^2}{2a^2}$ along the diagonal, outward.
Forces from the two $q$ charges (distance $a$ , perpendicular) give a resultant $\dfrac{\sqrt{2}kQq}{a^2}$ along the same diagonal.
For zero net force $q$ must attract $Q$ (opposite sign): $\dfrac{Q^2}{2} = -\sqrt{2}Qq \Rightarrow \dfrac{Q}{q} = -2\sqrt{2}$ .
So the answer is (d) $-2\sqrt{2}$ .

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