Explanation

(b) The charge $A\ $experiences three forces $F_{1},F_{2}$ and $F_{3}$ as shown in figure.
[image:1_image413.emf]
Now
$F_{1} = \frac{1}{4\pi\varepsilon_{0}\ }\frac{q_{1}q_{2}}{a^{2}}$ (along $AB$)
$F_{2} = \frac{1}{4\pi\varepsilon_{0}\ }\frac{q_{1}q_{2}}{a^{2}}$ (along $AD$)
$F_{3} = \frac{1}{4\pi\varepsilon_{0}\ }\frac{q_{1}^{2}}{a^{2}}$
The resultant of $F_{1}$ and $F_{2}$ should be equal and opposite to $F_{3}$ to keep the system in equilibrium.
Resultant $F_{1}$ and $F_{2}$=$F_{R} = \sqrt{F_{1}^{2} + \ F_{2}^{2}}$
$= \frac{1}{4\pi\ }\frac{q_{1}q_{1}}{\varepsilon_{0}a^{2}}\sqrt{2}$
For equilibrium,
$\frac{1}{4\pi\varepsilon_{0}}\frac{q_{1}q_{2}}{a^{2}}\sqrt{2} = \frac{1}{4\pi\varepsilon_{0}}\frac{q_{1}^{2}}{a^{2}}$
$\frac{q_{1}}{q_{2}} = \sqrt{2}$