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 the $$Q/q$$ equals

Consider a square of side $$a$$. Label the corners in order as $$A(0,0),\;B(a,0),\;C(a,a),\;D(0,a)$$.

Charges placed:
  • $$Q$$ at the opposite corners $$A$$ and $$C$$.
  • $$q$$ at the remaining corners $$B$$ and $$D$$.

We need the net electrostatic force on the charge $$Q$$ at corner $$A$$ to be zero.

Forces on $$Q$$ at $$A$$

1. Force due to the other $$Q$$ at $$C$$ (diagonal corner)
  Distance $$AC = a\sqrt{2}$$.
  Magnitude $$F_{AC} = k\dfrac{Q^2}{(a\sqrt{2})^{\,2}} = \dfrac{kQ^2}{2a^{2}}$$.
  Direction: from $$A$$ towards $$C$$ makes $$45^{\circ}$$ with both axes; because both charges $$Q$$ are alike, the force on $$A$$ is repulsive, i.e. along $$-\hat i-\hat j$$.
  Components: $$F_{AC,x} = F_{AC,y} = -\dfrac{kQ^2}{2a^{2}}\dfrac{1}{\sqrt{2}} = -\dfrac{kQ^2}{2\sqrt{2}\,a^{2}}$$.

2. Force due to $$q$$ at $$B(a,0)$$ (adjacent corner on the x-axis)
  Distance $$AB = a$$.
  Magnitude $$F_{AB} = k\dfrac{|Qq|}{a^{2}}$$.
  Direction: along the x-axis. Writing the x-component with sign included,
$$F_{AB,x}= -\dfrac{kQq}{a^{2}}$$ (positive if $$q$$ is opposite in sign to $$Q$$, negative if same sign).

3. Force due to $$q$$ at $$D(0,a)$$ (adjacent corner on the y-axis) is analogous:
  $$F_{AD,y}= -\dfrac{kQq}{a^{2}}$$.

Setting the net force to zero

Total x-component:
$$F_x = F_{AB,x} + F_{AC,x}= -\dfrac{kQq}{a^{2}}-\dfrac{kQ^2}{2\sqrt{2}\,a^{2}}=0$$.

Total y-component gives the same equation, so one condition suffices:

$$-\dfrac{kQq}{a^{2}}-\dfrac{kQ^2}{2\sqrt{2}\,a^{2}}=0$$

Cancel the common factor $$\dfrac{k}{a^{2}}$$:

$$-Qq-\dfrac{Q^{2}}{2\sqrt{2}}=0$$

$$Qq = -\dfrac{Q^{2}}{2\sqrt{2}}$$

Divide by $$Q\;(Q\neq0)$$:

$$q = -\dfrac{Q}{2\sqrt{2}}$$

Hence

$$\dfrac{Q}{q}= -2\sqrt{2}$$

which matches Option A.

Therefore, the required ratio is:
Option A which is: $$-2\sqrt{2}$$