Electrostatic energy increase on equal-drop coalescence ($V_f=2^{2/3}{V}_i,U_f=2^{2/3}{U}_i$)

Two isolated conducting drops of radius $a$ and equal charge merge at fixed total charge and volume. The new radius is $2^{1/3}a$, so its capacitance grows by only $2^{1/3}$ while total charge doubles. Relative to the sum of the initially separated self energies, the final electric field energy grows by $2^{2/3}$. Decreased surface energy and/or applied mechanical work can supply that increase. Sufficiently large like charges need not coalesce spontaneously; energy conservation is not a guarantee of merging. A maintained external voltage would define a different problem.