The radius of a soap bubble is \(r\) and the surface tension of the soap solution is \(S\). The electric…
The radius of a soap bubble is \(r\) and the surface tension of the soap solution is \(S\). The electric potential to which the soap bubble be raised by charging it so that the pressure inside the bubble becomes equal to the pressure outside the bubble is ( \(\varepsilon_0=\) permittivity of the free space)
\(\sqrt{\frac{\mathrm{Sr}}{8 \varepsilon_0}}\)
\(\sqrt{\frac{\mathrm{Sr}}{4 \varepsilon_0}}\)
\(\sqrt{\frac{4 \mathrm{~S} r}{\varepsilon_0}}\)
\(\sqrt{\frac{8 \mathrm{~S} r}{\varepsilon_0}}\)
Solution
\(\because\) Pressure due to surface tension inside the
\(\text {soap bubble, } \quad p_i=\frac{4 S}{r}\) ...(i)
Electrostatic pressure outside the soap bubble,
\(p_0=\frac{\sigma^2}{2 \varepsilon_0}\) ...(ii)
where, \(\sigma=\) surface charge density and \(\varepsilon_0=\) permittivity of the free space
\(\because\) Electric potential, \(V=\frac{k Q}{r}\)
\(\Rightarrow \quad V=\frac{k(\sigma A)}{r} \quad(\because Q=\sigma . A)\)
\(\because\) Area of sphere, \(A=4 \pi r^2\)
So, \(V=\frac{1}{4 \pi \varepsilon_0}\left(\frac{\sigma 4 \pi r^2}{r}\right)=\frac{\sigma r}{\varepsilon_0} \quad\left(\because k=\frac{1}{4 \pi \varepsilon_0}\right)\) or
\(\sigma=\frac{\varepsilon_0 V}{r}\)
According to the question,
\(p_i=p_0\)
From Eqs. (i) and (ii), we get
\(\frac{4 S}{r}=\frac{\sigma^2}{2 \varepsilon_0} \Rightarrow \frac{4 S}{r}=\frac{\left(\frac{\varepsilon_0 V}{r}\right)^2}{2 \varepsilon_0} \Rightarrow V=\sqrt{\frac{8 S r}{\varepsilon_0}}\)
Hence, the pressure outside the bubble is \(\frac{\sqrt{8 S r}}{\varepsilon_0}\).