The radius of a soap bubble is r and the surf | vedclass

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(D) The excess pressure inside a soap bubble due to surface tension is given by $p_i = \frac{4S}{r}$.When the bubble is charged,an outward electrostat

Vedclass · Accepted Answer

$\sqrt{\frac{8 S r}{\varepsilon_0}}$

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(D) The excess pressure inside a soap bubble due to surface tension is given by $p_i = \frac{4S}{r}$.When the bubble is charged,an outward electrostatic pressure is exerted on its surface,given by $p_e = \frac{\sigma^2}{2\varepsilon_0}$,where $\sigma$ is the surface charge density.The electric potential $V$ of a charged spherical bubble is $V = \frac{Q}{4\pi\varepsilon_0 r}$. Since $Q = \sigma(4\pi r^2)$,we have $V = \frac{\sigma r}{\varepsilon_0}$,which implies $\sigma = \frac{\varepsilon_0 V}{r}$.Substituting $\sigma$ into the electrostatic pressure formula: $p_e = \frac{(\varepsilon_0 V / r)^2}{2\varepsilon_0} = \frac{\varepsilon_0 V^2}{2r^2}$.For the pressure inside to equal the pressure outside,the excess pressure due to surface tension must be balanced by the electrostatic pressure: $\frac{4S}{r} = \frac{\varepsilon_0 V^2}{2r^2}$.Solving for $V$: $V^2 = \frac{8Sr}{\varepsilon_0} \Rightarrow V = \sqrt{\frac{8Sr}{\varepsilon_0}}$.

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