What happens when a soap bubble is charged?

Two drops of equal radius coalesce to form a bigger drop. What is the ratio of the surface energy of the bigger drop to that of a smaller one?

Consider a water droplet of diameter $0.2 \,mm$ where the outside pressure is $1.5 \,N / cm^2$ at $25^{\circ} C$. The pressure inside the droplet, when the surface tension at $25^{\circ} C$ is $0.08 \,N / m$ is

One end of a uniform glass capillary tube of radius $r = 0.025 \ cm$ is immersed vertically in water to a depth $h = 1 \ cm$. The excess pressure in $N/m^2$ required to blow an air bubble out of the tube is: (Surface tension of water $T = 7 \times 10^{-2} \ N/m$,Density of water $\rho = 10^3 \ kg/m^3$,Acceleration due to gravity $g = 10 \ m/s^2$)

Two soap bubbles of radii $r_1$ and $r_2$ coalesce in vacuum under isothermal conditions to form a bigger bubble of radius $R$. What is the radius of the bigger bubble?

$A$ liquid drop of radius $R$ is broken into $n$ identical small droplets. The work done is $[T = \text{surface tension of the liquid}]$