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 $({\epsilon }_0=$ permittivity of the free space $)$

• A. $\sqrt{\frac{Sr}{8{\in }_0}}$
• B. $\sqrt{\frac{Sr}{4{\in }_0}}$
• C. $\sqrt{\frac{4Sr}{{\in }_0}}$
• D. $\sqrt{\frac{8Sr}{{\in }_0}}$

Answer 1

Answer: (D)

$\because$ Pressure due to surface tension inside the
soap bubble. $p_i=\frac{4S}{r}\dots (i)$
Electrostatic pressure outside the soap bubble,
$p_0=\frac{{\sigma }^2}{2{\epsilon }_0}\dots (ii)$
where, $\sigma =$ surface charge density and ${\epsilon }_0=$ permittivity of the free space
$\because$ Electric potential, $V=\frac{kQ}{r}$
$\Rightarrow V=\frac{k(\sigma A)}{r}$
$(\because Q=\sigma .A)$
$\because$ Area of sphere, $A=4\pi r^2$
So, $V=\frac{1}{4\pi {\epsilon }_0}\left(\frac{\sigma 4\pi r^2}{r}\right)=\frac{\sigma r}{{\epsilon }_0}$
$\left(\because k=\frac{1}{4\pi {\epsilon }_0}\right)$
or $=\frac{{\epsilon }_{0}V}{r}$
According to the question,
$p_i=p_0$
From Eqs. (i) and (ii), we get
$\frac{4S}{r}=\frac{{\sigma }^2}{2{\epsilon }_0}$
$\Rightarrow \frac{4S}{r}=\frac{{\left(\frac{{\epsilon }_{0}V}{r}\right)}^2}{2{\epsilon }_0}$
$\Rightarrow V=\sqrt{\frac{8Sr}{{\epsilon }_0}}$
Hence, the pressure outside the bubble is $\frac{\sqrt{8Sr}}{{\epsilon }_0}$.