Four point charges, each of +q , are rigidly fixed at the four...

$F_1=$ Net electrostatic force on anyone charge due to rest of three charges =$\frac{1}{4 \pi \varepsilon_0} \frac{q^2}{a^2}$\left(\sqrt{2}+\frac{1}{2}\right)F_2=$Surface tension force$=\gammaa$If we see the equilibrium of line$B C$, then,$\quad 2 F_1 \cos 45^{{}^{\circ}}=F_2$or$\quad \sqrt{2}F_1=F_2$or$\frac{1}{4 \pi \varepsilon_0} \frac{q^2}{a^2}\left(2+\frac{1}{\sqrt{2}}\right)=\gammaa\therefore \quad a^3=\frac{1}{4 \pi \varepsilon_0}\left(2+\frac{1}{\sqrt{2}}\right) \frac{q^2}{\gamma}$or$\quad a=\left\{\frac{1}{4 \pi \varepsilon_0}\left(2+\frac{1}{\sqrt{2}}\right)\right\}^{1 / 3}\left[\frac{q^2}{\gamma}\right]^{1 / 3}=k\left[\frac{q^2}{\gamma}\right]^{1 / 3}$where,$\quad k=\left\{\frac{1}{4 \pi \varepsilon_0}\left(2+\frac{1}{\sqrt{2}}\right)\right\}^{1 / 3}$Therefore,$N=3 Answer is 3 . Analysis of Question (i) Question is moderately difficult. (ii) In my opinion problems of surface tension or viscosity are always not very simple questions.