Surface charge density of soap bubble of radius ‘r’ and surface tension T is $σ$ . If P is excess pressure, the value of $σ$ is

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${[4 \in_{0} (\frac{2 T}{r} - P)]}^{1 / 2}$

$\frac{4 T}{r}$

$\in_{0} {[\frac{4 T}{r} - P]}^{3 / 2}$

${[2 \in_{0} (\frac{4 T}{r} - P)]}^{1 / 2}$

answer is B.

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Detailed Solution

Electrostatic pressure due to surface charge density is

$\frac{σ^{2}}{2 ε_{0}}$

$So excess pressure P + \frac{σ^{2}}{2 ε_{0}} = \frac{4 T}{r}$

$\frac{σ^{2}}{2 ε_{0}} = \frac{4 T}{r} - P σ = {[2 ε_{0} (\frac{4 T}{r} - P)]}^{1 / 2}$

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