Question

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 $a=c{\left[\frac{q^2}{\gamma }\right]}^{1/N}$, where '$c$' is a constant. Then $N$ is

Answer 1

From figure,
Balancing vertical forces, we get
$2\gamma a=\frac{2k{q}^2}{a^2}+2\left(\frac{k{q}^2}{2a^2}\cos {45}^{\circ }\right)$
$\gamma a=\frac{k{q}^2}{a^2}(1+\frac{1}{2\sqrt{2}})$
$\Rightarrow a={\left(\frac{q^2}{\gamma }\right)}^{1/3}{\left[(1+\frac{1}{2\sqrt{2}})k\right]}^{1/3}=c{\left[\frac{q^2}{\gamma }\right]}^{1/N}$
On comparing with the relation, we get $N=3$