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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
$3$
$6$
$4$
$5$
Solution
$F_1=$ Net electrostatic force on any one 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 this equation we get $a =2$.
If we see the equilibrium of line $B C$, then .
$2 F_1 \cos 45^{\circ}=F_2 \text { or } \sqrt{2} F_1=F_2 \text { or } \frac{1}{4 \pi \varepsilon_0} \frac{q^2}{a^2}(2 +\frac{1}{\sqrt{2}}=\gamma a$
$\therefore a^3=\frac{1}{4 \pi \varepsilon_0}\left(2+\frac{1}{\sqrt{2}} \frac{q^2}{\gamma} \text { or } a\right.$
$=\left\{\frac{1}{4 \pi \varepsilon_0}\left(2+\frac{1}{\sqrt{2}}\right\}^{1 / 3}\left[\frac{q^2}{\gamma}\right)\right]^{2 / k}=k\left[\frac{q^2}{\gamma}\right]^{1 / 3}$
Where $k=\left\{\frac{1}{4 \pi \varepsilon_0}\left(2+\frac{1}{2}\right)\right\}^{1 / 3}$ therefore $N =3$
