Question

A soap bubble of radius $3 \,cm$ is charged with $9\, nC$. Find the excess pressure inside the bubble (in SI units). Consider surface tension of soap solution $=3 \times 10^{-3} N m ^{-1}$.

Answer 1

Excess pressure due to surface tension $=\frac{4 T }{ r }$.
(This pressure tends to contract the bubble.)
Outward pull due to electric charge per unit area $=\frac{\sigma^{2}}{2 \varepsilon_{0}}$
$\therefore $ Excess pressure inside a charged soap bubble
$=\frac{4 T }{ r }-\frac{\sigma^{2}}{2 \varepsilon_{0}}$
Now, $q$ (total charge) $=4 \pi r^{2} \sigma$
$\therefore \sigma=\frac{ q }{4 \pi r ^{2}}$
$\therefore $ Excess pressure
$\Delta P =\frac{4 T }{ r }-\frac{ q ^{2}}{32 \pi^{2} r ^{4} \varepsilon_{0}} $
$=\frac{4 \times 3 \times 10^{-3}}{3 \times 10^{-2}}-\frac{ q ^{2} \varepsilon_{0}}{2 \times 4 \pi \varepsilon_{0} \times 4 \pi \varepsilon_{0} \times r ^{4}} $
$=0.4-\frac{\left(9 \times 10^{-9}\right)^{2} \times\left(9 \times 10^{9}\right)^{2} \times 8.85 \times 10^{-12}}{2 \times\left(3 \times 10^{-2}\right)^{4}}$
$=0.4-\frac{9 \times 9 \times 8.85 \times 10^{-12}}{2 \times 10^{-8}}$
$=0.4-\frac{716.85}{2} \times 10^{-4}$
$=0.4-0.036$
$\approx 0.36\, N m ^{-2}$