Physics - Electrostatics Question with Solution | TestHub

The charge$\mathrm{A}$experiences three forces${\mathrm{F}}_1$,${\mathrm{F}}_2$and${\mathrm{F}}_3$as shown in the figure.



Now,${\mathrm{F}}_1=\frac{1}{4{\mathrm{\pi \epsilon }}_0\mathrm{ }}\frac{{\mathrm{q}}_1{\mathrm{q}}_2}{{\mathrm{a}}^2}$(along$\mathrm{AB}$)

${\mathrm{F}}_2=\frac{1}{4{\mathrm{\pi \epsilon }}_0\mathrm{ }}\frac{{\mathrm{q}}_1{\mathrm{q}}_2}{{\mathrm{a}}^2}$(along$\mathrm{AD}$)

${\mathrm{F}}_3=\frac{1}{4{\mathrm{\pi \epsilon }}_0\mathrm{ }}\frac{{\mathrm{q}}_1^2}{{\mathrm{a}}^2}$

The resultant of${\mathrm{F}}_1$and${\mathrm{F}}_2$should be equal and opposite to${\mathrm{F}}_3$to keep the system in equilibrium.

Resultant of$F_1$and${\mathrm{F}}_2$$={\mathrm{F}}_{\mathrm{R}}=\sqrt{{\mathrm{F}}_1^2+\mathrm{ }{\mathrm{F}}_2^2}$

$=\frac{1}{4\mathrm{\pi }\mathrm{ }}\frac{{\mathrm{q}}_1{\mathrm{q}}_1}{{\mathrm{\epsilon }}_0{\mathrm{a}}^2}\sqrt{2}$

For equilibrium,

$\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{{\mathrm{q}}_1{\mathrm{q}}_2}{{\mathrm{a}}^2}\sqrt{2}=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{{\mathrm{q}}_1^2}{{\mathrm{a}}^2}$

$\frac{{\mathrm{q}}_1}{{\mathrm{q}}_2}=\sqrt{2}$