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(C) The excess pressure inside a soap bubble due to surface tension is given by $p = \frac{4S}{R}$,where $S$ is the surface tension and $R$ is the rad

Vedclass · Accepted Answer

$11.3 	imes 10^{-4} \ N \ m^{-1}$

Vedclass · Answer

(C) The excess pressure inside a soap bubble due to surface tension is given by $p = \frac{4S}{R}$,where $S$ is the surface tension and $R$ is the radius.Given that the electrostatic pressure is $p = \frac{\sigma^2}{2 \varepsilon_0}$.According to the problem,the excess pressure due to surface tension is equal to the electrostatic pressure:$\frac{4S}{R} = \frac{\sigma^2}{2 \varepsilon_0}$Rearranging for $S$:$S = \frac{\sigma^2 R}{8 \varepsilon_0} \quad \dots (i)$Given values: $\sigma = 20 \ \mu C \ m^{-2} = 20 	imes 10^{-6} \ C \ m^{-2}$,$R = 0.2 \ mm = 0.2 	imes 10^{-3} \ m$,and $\varepsilon_0 = 8.85 	imes 10^{-12} \ C^2 \ N^{-1} \ m^{-2}$.Substituting these values into equation $(i)$:$S = \frac{(20 	imes 10^{-6})^2 	imes (0.2 	imes 10^{-3})}{8 	imes 8.85 	imes 10^{-12}}$$S = \frac{400 	imes 10^{-12} 	imes 0.2 	imes 10^{-3}}{70.8 	imes 10^{-12}}$$S = \frac{80 	imes 10^{-15}}{70.8 	imes 10^{-12}} \approx 1.13 	imes 10^{-3} \ N \ m^{-1} = 11.3 	imes 10^{-4} \ N \ m^{-1}$.

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