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 $P_0=10^{2}Pa,T=0.07N{m}^{-1},r=1cm,{ϵ}_0=9\times 10^{-12}{C}^{2}N{m}^{-2}.$

Answer 1

Answer: 7

Recall the ideal gas equation in terms of pressure and volume,
$PV=nRT$ , so it the temperature is constant then,
$P_1{V}_1=P_2{V}_2$
$\left(P_0+\frac{4T}{r}\right)\frac{4}{3}\pi r^3=\left(P_0+\frac{4T}{2r}-\frac{{\sigma }^2}{2{\epsilon }_0}\right)\frac{4}{3}\pi 8r^3$
$P_0+\frac{4T}{r}=8P_0+16\frac{T}{r}-\frac{4{\sigma }^2}{{ϵ}_0}$
$\frac{4{\sigma }^2}{{ϵ}_0}=7P_0+\frac{12T}{r}$
$\frac{4{\sigma }^2}{9\times 10^{-12}}=700+84$
$4{\sigma }^2=784\times 9\times 10^{-12}$
$2\sigma =28\times 3\times 10^{-6}$
$\sigma =14\times 3\times 10^{-6}$
$\frac{Q}{4\pi \times 4\times 10^{-4}}=14\times 3\times 10^{-6}$
$\frac{Q}{96\pi \times 10^{-10}}=\frac{14\times 3\times 10^{-6}}{6\times 10^{-6}}=7$