Capacitors and Kirchoff's Voltage Law If we connect capacitor to voltage source, its voltage will be equal to voltage of the source when capacitor is fully charged due to Kirchoff's voltage law and no current will flow in a circuit any longer.
If we had a theorethical capacitor with no or very little capacitance than almost no charge would develop on it for certain voltage.
If so, how can voltage on capacitor develop if there is no or almost no charge separated on the plates since voltage applied can be arbitrarily big?
If we for example connected such capacitor to some high voltage like 50 kV, how can such voltage develop on capacitor with no or almost no charge separated on the plates of the capacitor?
 A: If we had a theoretical capacitor with no or very little capacitance than almost no charge would develop on it for certain voltage
This statement is misleading in that there is a capacitance in a circuit even when think there is none.
For example consider a series circuit consisting of a battery, a resistor and a switch which is open.
Is there a capacitor in the circuit?
Indeed there is even though its capacitance is so small that you will probably never notice its effect.
The switch acts as a capacitor with a very small capacitance and when open the potential difference across it is equal to the emf of the battery and so there are charges stored on the two parts of the open switch even though the amount of charge is very, very small.
So even though $C$ may be very, very small, $Q=CV$ still applies.
A: Just consider a spherical capacitor that is charged with a single electron $q=e$ (i.e. the smallest possible charge) against infinity (suppose the electron charge could be distributed evenly over the sphere). Then the spherically symmetric field outside the sphere is identical to the field of a point-like localized electron, and thus, also the potential. Since the potential of a point charge has an $1/r$ behavior, the voltage of the capacitor, i.e. the potential on the sphere's surface (of radius $R$) also has a $1/R$ dependency. So
$$U=\frac{ke}{R}$$
which means, there is no limit to the voltage that a spherical capacitor can carry, just because there is no limit to the value of the potential of a point charge at distance $R$. The closer to the center you get (the smaller the radius of the sphere), the higher the voltage. Even with a single electron you can achieve arbitrarily high potentials (and hence, voltages).
Your question wrongly assumes that there is a kind of natural limit to the voltage of a given amount of charge. There is not.
