Question

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 the charge on the drop is?
(A) $2\sqrt{\dfrac{2rT}{{{\varepsilon }_{0}}}}$
(B) $8\pi r\sqrt{2rT{{\varepsilon }_{0}}}$
(C) $8\pi r\sqrt{rT{{\varepsilon }_{0}}}$
(D) $8\pi r\sqrt{\dfrac{2rT}{{{\varepsilon }_{0}}}}$

Answer 1

The charge is isolated and so it will produce an electric field around itself and it will extend to a large region depending upon the magnitude of the charge. The radius of the soap bubble is mentioned. The surface tension we know is defined as the tendency of the liquid surfaces to shrink as to minimize the surface area and as such the radius tends to decrease.

Complete step by step answer:
Now inside the soap bubble the pressure must be greater than $\dfrac{4T}{r}$ the outside pressure for the bubble to exist. Now this excess pressure will come at the expense of some external agent and that agent here is the charge. So, the excess pressure here is provided by the charge on the bubble.
$\dfrac{4T}{r}=\dfrac{{{\rho }^{2}}}{2{{\varepsilon }_{0}}}$
where $\rho$ is the charge density of the soap bubble, ${{\varepsilon }_{0}}$ is the permittivity of the free space, ‘T’ is the surface tension of soap solution.
But we know that the surface is spherical and so charge density in terms of charge can be written as: $\rho =\dfrac{Q}{4\pi {{r}^{2}}}$, where r is the radius and Q is the charge. Thus,
$\dfrac{4T}{r}=\dfrac{{{\rho }^{2}}}{2{{\varepsilon }_{0}}}$
$\Rightarrow \dfrac{4T}{r}=\dfrac{{{Q}^{2}}}{2{{\varepsilon }_{0}}\times 16{{\pi }^{2}}{{r}^{2}}}$
$\Rightarrow {{Q}^{2}}=\dfrac{4\times 2\times 16T{{\varepsilon }_{0}}\times {{\pi }^{2}}{{r}^{2}}}{r}$
$\therefore Q=8\pi r\sqrt{2rT{{\varepsilon }_{0}}}$

So, the correct option comes out to be B.

Note: The origin of the surface tension lies within the cohesive forces between liquid molecules which are responsible for the phenomenon of surface tension. The molecules at the surface do not have other like molecules on all sides of them and thus they cohere more strongly to those directly associated with them on the surface. Because of this a surface "film" which makes it more difficult to move an object through the surface than to move it when it is completely submersed.