= Electrostatic energy increase on equal-drop coalescence
{title2=$V_f=2^{2/3}V_i,\quad U_f=2^{2/3}U_i$}
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.
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