ELECTROSTATICS BYU 54

Solution: Question 54

Solution to Question 54

Physics Principle: Small Oscillations about Equilibrium.

1. Equilibrium Condition:
The bubble is stable when the inward force due to surface tension balances the outward electrostatic pressure.
Potential Energy \(U(r) = U_{surf} + U_{elec} = 8\pi\sigma r^2 + \frac{Q^2}{8\pi\epsilon_0 r}\).
At equilibrium radius \(R\), \(dU/dr = 0\): $$ 16\pi\sigma R – \frac{Q^2}{8\pi\epsilon_0 R^2} = 0 \implies 16\pi\sigma = \frac{Q^2}{8\pi\epsilon_0 R^3} $$

2. Restoring Force Constant (\(k_{eff}\)):
For small radial oscillations \(x\) where \(r = R+x\), the effective spring constant is the second derivative of \(U\) at \(R\). $$ k_{eff} = \frac{d^2U}{dr^2}\Bigg|_{r=R} = 16\pi\sigma + \frac{2 Q^2}{8\pi\epsilon_0 R^3} $$ Substitute the equilibrium condition term \(\frac{Q^2}{8\pi\epsilon_0 R^3} = 16\pi\sigma\): $$ k_{eff} = 16\pi\sigma + 2(16\pi\sigma) = 48\pi\sigma $$

3. Time Period:
Considering the mass \(m\) of the bubble film oscillating radially: $$ T = 2\pi \sqrt{\frac{m}{k_{eff}}} = 2\pi \sqrt{\frac{m}{48\pi\sigma}} $$ Simplifying: $$ T = \frac{2\pi}{4} \sqrt{\frac{m}{3\pi\sigma}} = \frac{\pi}{2} \sqrt{\frac{m}{3\pi\sigma}} = \sqrt{\frac{\pi^2 m}{12\pi\sigma}} $$ $$ T = \sqrt{\frac{\pi m}{12\sigma}} $$

Time Period: $$ T = \sqrt{\frac{\pi m}{12\sigma}} $$