Question

The radius of a soap bubble is $r$ and the surface tension of the soap solution is $S$. The electric potential to which the soap bubble be raised by charging it so that the pressure inside the bubble becomes equal to the pressure outside the bubble is $\left(\varepsilon_{0}=\right.$ permittivity of the free space $)$

• A. $\sqrt{\frac{Sr}{8\in_{0}}}$
• B. $\sqrt{\frac{Sr}{4 \in_{0}}}$
• C. $\sqrt{\frac{4 Sr}{\in_{0}}}$
• D. $\sqrt{\frac{8 Sr}{\in_{0}}}$

Answer 1

Answer: (D)

$ \because$ Pressure due to surface tension inside the
soap bubble. $p_{i}=\frac{4 S}{r}\,\,\,\,\,\,\,\,\,\dots(i)$
Electrostatic pressure outside the soap bubble,
$p_{0}=\frac{\sigma^{2}}{2 \varepsilon_{0}}\,\,\,\,\,\,\,\,\dots(ii)$
where, $\sigma=$ surface charge density and $\varepsilon_{0}=$ permittivity of the free space
$\because$ Electric potential, $V =\frac{k Q}{r} $
$\Rightarrow V=\frac{k(\sigma A)}{r}$
$(\because Q=\sigma . A)$
$\because$ Area of sphere, $ A=4 \pi r^{2}$
So, $V=\frac{1}{4 \pi \varepsilon_{0}}\left(\frac{\sigma 4 \pi r^{2}}{r}\right)=\frac{\sigma r}{\varepsilon_{0}}$
$\left(\because k=\frac{1}{4 \pi \varepsilon_{0}}\right) $
or $=\frac{\varepsilon_{0} V}{r}$
According to the question,
$p_{i}=p_{0}$
From Eqs. (i) and (ii), we get
$\frac{4 S}{r}=\frac{\sigma^{2}}{2 \varepsilon_{0}}$
$ \Rightarrow \frac{4 S}{r}=\frac{\left(\frac{\varepsilon_{0} V}{r}\right)^{2}}{2 \varepsilon_{0}} $
$\Rightarrow V=\sqrt{\frac{8 S r}{\varepsilon_{0}}}$
Hence, the pressure outside the bubble is $\frac{\sqrt{8 S r}}{\varepsilon_{0}}$.