A soap bubble of radius R and surface tension | vedclass

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Question

(A) Initially,the pressure inside the soap bubble is $P_{	ext{in}} = \frac{4T}{R}$.When the bubble is charged,the electrostatic pressure (outward) is

Vedclass · Accepted Answer

$Q = \sqrt {768{\pi ^2}{R^3}{\epsilon _0}T} $

Vedclass · Answer

(A) Initially,the pressure inside the soap bubble is $P_{	ext{in}} = \frac{4T}{R}$.When the bubble is charged,the electrostatic pressure (outward) is $P_e = \frac{\sigma^2}{2\epsilon_0}$.For a soap bubble in a vacuum,the pressure balance at the final radius $2R$ is given by:$P_{	ext{in}}' + P_e = \frac{4T}{2R}$,where $P_{	ext{in}}'$ is the new internal pressure.Assuming the process is isothermal,$P_{	ext{in}}' = P_{	ext{in}} \left( \frac{V_i}{V_f} ight) = \frac{4T}{R} \left( \frac{\frac{4}{3}\pi R^3}{\frac{4}{3}\pi (2R)^3} ight) = \frac{4T}{R} \cdot \frac{1}{8} = \frac{T}{2R}$.Substituting this into the balance equation:$\frac{T}{2R} + \frac{\sigma^2}{2\epsilon_0} = \frac{2T}{R}$$\frac{\sigma^2}{2\epsilon_0} = \frac{2T}{R} - \frac{T}{2R} = \frac{3T}{2R}$$\sigma^2 = \frac{3T\epsilon_0}{R} \implies \sigma = \sqrt{\frac{3T\epsilon_0}{R}}$.The total charge $Q$ is $\sigma \cdot A$,where $A = 4\pi(2R)^2 = 16\pi R^2$.$Q = \sqrt{\frac{3T\epsilon_0}{R}} \cdot 16\pi R^2 = \sqrt{\frac{3T\epsilon_0}{R} \cdot 256\pi^2 R^4} = \sqrt{768\pi^2 R^3 \epsilon_0 T}$.

$A$ soap bubble of radius $R$ and surface tension $T$ is formed in a vacuum. It is slowly charged so that it slowly expands. It stops charging when the radius becomes $2R$. Find the amount of charge given to the bubble.

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