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(\left[\right. \frac{q^{2}}{\gamma} \left]\right.\right)^{1 / N}, where 'k' is a constant. Then N is

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Physics

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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 $γ$ . The system of charges and planar film are in equilibrium, and $a = k {[\frac{q^{2}}{γ}]}^{1 / N}$ , where 'k' is a constant. Then N is

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$F_{AC} = \frac{q^{2}}{8 {πε}_{0} a^{2}}$

$F_{AD} = F_{AB} = \frac{q^{2}}{4 {πε}_{0} a^{2}}$

$F_{R} = \frac{q^{2}}{4 {πε}_{0} a^{2}} (2 \cos 45 ° + \frac{1}{2})$

$= r (2) (BD) = 2 r (\sqrt{2} a)$

$\Rightarrow a^{3} = \frac{q^{2} (\sqrt{2} + \frac{1}{2})}{8 \sqrt{2} {πε}_{0} r}$

$a = k {(\frac{q^{2}}{r})}^{1 / 3} \Rightarrow N = 3$

Where $k = {(\frac{\sqrt{2} + \frac{1}{2}}{8 \sqrt{2} π})}^{1 / 3}$

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