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Q.

An isolated and charged spherical soap bubble has a radius r and the pressure inside as well as outside is atmospheric. If T is the surface tension of soap solution, then charge on drop is

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a

$2 \sqrt{\frac{2 r T}{\in_{0}}}$

b

$8 π r \sqrt{2 r T \in_{0}}$

c

$8 π r \sqrt{r T \in_{0}}$

d

$8 π r \sqrt{\frac{2 r T}{\in_{0}}}$

answer is B.

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Detailed Solution

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Inside pressure must be $\frac{4 T}{r}$ greater than outside pressure in bubble. This excess pressure is provided by charge on bubble .So, we have
Question Image
$\frac{4 T}{r} = \frac{σ 2}{2 \in_{0}}$
Since $σ = \frac{Q}{4 π r^{2}}$
$\Rightarrow \frac{4 T}{r} = \frac{Q^{2}}{16 π^{2} r^{4} \times 2 \in_{0}}$
$\Rightarrow Q = 8 π r \sqrt{2 r T \in_{0}}$

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