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.…

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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

Aicharya · Accepted Answer

$F_1=$ 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. !

Four point charges, each of $+q$, are rigidly fixed at the four corners of a square planar soap film of side…

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