question_answer

An isolated and charged spherical soap bubble has a radius 'r' and the pressure inside is atmospheric. If 'T' is the surface tension of soap solution, then charge on drop is:

A) \[2\sqrt{\frac{2\,\,r\,\,T}{{{\varepsilon }_{0}}}}\]

B) \[8\,\,\pi \,\,r\,\,\sqrt{2\,\,r\,\,T\,\,{{\varepsilon }_{0}}}\]

C) \[8\,\,\pi \,\,r\,\,\sqrt{r\,\,T\,\,{{\varepsilon }_{0}}}\]

D) \[8\,\,\pi \,\,r\,\,\sqrt{\frac{2\,\,r\,\,T}{{{\varepsilon }_{0}}}\,\,}\]

Correct Answer: B

Solution :

Inside pressure must be \[\frac{4T}{r}\] greater than outside pressure in bubble. This excess pressure is provided by charge on bubble.

\[\frac{4T}{r}=\frac{{{\sigma }^{2}}}{2{{\varepsilon }_{0}}}\]

\[\frac{4T}{r}=\frac{{{Q}^{2}}}{16{{\pi }^{2}}{{r}^{4}}\times 2{{\varepsilon }_{0}}}......\left[ \sigma =\frac{Q}{4\pi {{r}^{2}}} \right]\]

\[Q=8\pi r\sqrt{2rT{{\varepsilon }_{0}}}\]