A soap bubble (surface tension $= T$ ) is charged to a maximum surface density of charge $= σ .$ When it is just going to burst? Its radius R is given by

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$R = \frac{σ^{2}}{8 ε_{0} T}$

$R = 8 ε_{0} \frac{T}{σ^{2}}$

$R = \frac{σ}{\sqrt{8 ε_{0} T}}$

$R = \frac{\sqrt{8 ε_{0} T}}{σ}$

answer is B.

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Detailed Solution

The pressure due to surface tension $= \frac{4 T}{R}$
The pressure due to electrostatic forces $= \frac{σ^{2}}{2 ε_{0}}$
Just before the bubble bursts.

$\frac{4 T}{R} = \frac{σ^{2}}{2 ε_{0}} o r R = \frac{8 T ε_{0}}{σ^{2}}$

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