When a soap bubble of radius $0.2 \mathrm{~mm}$ is charged, it experiences an outward electrostatic pressure…
When a soap bubble of radius $0.2 \mathrm{~mm}$ is charged, it experiences an outward electrostatic pressure of magnitude $\frac{\sigma^2}{2 \varepsilon_0}$, where $\sigma=20 \mu \mathrm{Cm}^{-2}$ is the surface charge density. If the excess pressure inside the soap bubble due to the surface tension is same as
this electrostatic pressure, then the surface tension of the soap solution is
$$
\left(\varepsilon_0=8.85 \times 10^{-12} \mathrm{C}^2 \mathrm{~N}^{-1} \mathrm{~m}^{-2}\right)
$$
$8.85 \times 10^{-4} \mathrm{Nm}^{-1}$
$12.4 \times 10^{-4} \mathrm{Nm}^{-1}$
$11.3 \times 10^{-4} \mathrm{Nm}^{-1}$
$90 \times 10^{-4} \mathrm{Nm}^{-1}$
Solution
Excess pressure inside soap bubble due to surface tension, $p=\left(\frac{4 S}{R}\right)$, where $S=$ surface tension and $R=$ radius.
$\Rightarrow$ Electrostatic pressure, $p=\frac{\sigma^2}{2 \varepsilon_0}$ According to the question,
$
\frac{4 S}{R}=\frac{\sigma^2}{2 \varepsilon_0}
$
Putting all values in Eq. (i), we get
$
\begin{aligned}
& S=\frac{\left(20 \times 10^{-6}\right)^2 \times\left(0.2 \times 10^{-3}\right)}{\left(8 \times 8.85 \times 10^{-12}\right)} \\
& \Rightarrow \quad S=11.3 \times 10^{-4} \mathrm{~N} / \mathrm{m}
\end{aligned}
$
