In Problem 10.4, we are given that $\sigma = 0.5 \, \text{J/m}^2$ and $\Delta G_{\text{v}} = -10^4 \, \text{J/m}^3$. Plugging these values into the equation, we get:
$$r_{\text{crit}} = \frac{2(0.5 \, \text{J/m}^2)}{-10^4 \, \text{J/m}^3} = -10^{-4} \, \text{m}
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