Four point charges, each of \(+q\), are rigidly fixed at the four corners of a square planar soap film of side \(a\). The surface tension of the soap film is \(\gamma\). The system of charges and planar film are in equilibrium, and \(a=k\left[\frac{q^2}{\gamma}\right]^{1 / N}\), where \(k\) is a constant. Then, \(N\) is

\(F_1=\) 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 \(=\gamma a\)
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 \(\frac{1}{4 \pi \varepsilon_0} \frac{q^2}{a^2}\left(2+\frac{1}{\sqrt{2}}\right)=\gamma a\)
\(\therefore \quad a^3=\frac{1}{4 \pi \varepsilon_0}\left(2+\frac{1}{\sqrt{2}}\right) \frac{q^2}{\gamma}\)
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.