Sure, if you but in series 2 two batteries of 1.2V, 2200mAh, you get a 2.4V, 2200mAh battery. Not 2.4V, 1100 mAh, indeed.
But F is not a unit similar to Ah. 1 F = 1 Coulomb per volt. (Hence the formula Cu=q) And 1 Amp = 1 Coulomb per second. So 1 Amp×hour (Ah) = 1 Coulomb per second × 3600 seconds = 3600 coulombs. So Ah and Coulombs are both units of the same thing (with a 3600 conversion ratio) And 1 Farad is that, per volt.
To be more accurate 1C=1 As = 1/3600 Ah 1 Farad = 1 C/v = (1/3600) Ah/v
So if you want to convert, in Farad, my 2× 1.2V, 2200 mAh battery example, what you have is two 1.2V, 2.2 Ah = 1.2V,7920 C = 1.2V,7920/1.2=6600 F (6600 F × 1.2V is indeed 7920 C=2200 mAh)
Which turns into 2.4 V, 2.2 Ah = 2.4V,7920 C = 2.4V,7920/2.4=3300 F.
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Said otherwise, back, to your example, 5v, 10μF = 50 μC = 13.9 nAh 10v, 5μF = 50 μC = 13.9 nAh
So what you have is "two 5v, 13.9 nAh capacitors in series give one 10v, 13.9 nAh". Exactly like batteries.
We know Q = CV and we know caps in series have the same Q Let the equivalent capacitance = Ceq Ceq = Q/V = Q/(V1+V2)= C1V1/(V1+V2) but V2 = C1V1/C2 V1 + V2 = V1(1+C1/C2) = V1(C2+C1)/C2 Q/V = C1V1/(V1+V2) = C1V1/[V1(C2+C1)/C2] = C1C2/(C1+C2) = Ceq Hope this helps.
voltages add in batteries, and they add in caps, or any series components, like resistors.
two caps each with 5 volts on them, in series become 10 volts.
the capacitance value in series uses the formula 1/C = 1/C₁ + 1/C₂. That is due to the physical nature of the caps. Look at two equal caps in series, the equiv is the same plate size and twice the spacing, which means half the value.