Question

The radius and surface tension of a spherical soap bubble be $r$ and $T$ respectively. Find the charge required which is to be given to the bubble so that its radius will become $2r$ in equilibrium. Here atmospheric pressure is $P_{0},$ temperature of the air inside the bubble remains constant & the charge is assumed to uniformly distributed on the surface of the bubble. Find the value of $\frac{Q}{96 \pi \times 10^{- 10}}$ (in $C$ ). Take $P_{0}=10^{2}Pa,T=0.07Nm^{- 1},r=1cm,\epsilon _{0}=9\times 10^{- 12}C^{2}Nm^{- 2}.$

Answer 1

Recall the ideal gas equation in terms of pressure and volume,
$PV=nRT$ , so it the temperature is constant then,
$P_{1}V_{1}=P_{2}V_{2}$
$\left(P_{0}+\frac{4 T}{r}\right) \frac{4}{3} \pi r^{3}=\left(P_{0}+\frac{4 T}{2 r}-\frac{\sigma^{2}}{2 \varepsilon_{0}}\right) \frac{4}{3} \pi 8 r^{3}$
$P_{0}+\frac{4 T}{r}=8P_{0}+16\frac{T}{r}-\frac{4 \sigma ^{2}}{\epsilon _{0}}$
$\frac{4 \sigma ^{2}}{\epsilon _{0}}=7P_{0}+\frac{12 T}{r}$
$\frac{4 \sigma ^{2}}{9 \times 10^{- 12}}=700+84$
$4\sigma ^{2}=784\times 9\times 10^{- 12}$
$2\sigma =28\times 3\times 10^{- 6}$
$\sigma =14\times 3\times 10^{- 6}$
$\frac{Q}{4 \pi \times 4 \times 10^{- 4}}=14\times 3\times 10^{- 6}$
$\frac{Q}{96 \pi \times 10^{- 10}}=\frac{14 \times 3 \times 10^{- 6}}{6 \times 10^{- 6}}=7$