• # 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) $8\,\pi \,r\sqrt{\frac{2\,r\,T}{{{\varepsilon }_{0}}}}$

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