The general equation relating capacitance to electric charge and voltage is
C = Q/V where C is capacitance, Q is the electric charge applied to the plate of the capacitor, and V is the voltage difference.
If capacitors are in parallel the voltage on each plate is the same as that of the power supply and Q is distributed across the capacitors proportionally. Solving the capacitance equation for Q gives
Q = CV, thus for two capacitors in parallel you would have
Q = Q1 + Q2 where Q is the total charge provided by the power supply. Substituting the capacitance equation gives
CV = C1V1 + C2V2. Because V is the same for V1 and V2 and equal to V from the power supply, this can be simplified to
C = C1 + C2. If the two capacitors are 6 microF and 2 microF, than the equivalent capacitance for those two would be 8 micro F.
If these are now in series with two other capcitors we would get:
Solving this equation for V gives
V = Q/C
According to conservation of energy, a circuit in which the capacitors are in series, the voltage stored on each capacitor has to add up to the total voltage supplied by the power supply.
V = V1 + V2 + V3 where the 1, 2, 3 refer to the voltage on each capacitor in the circuit, and V the voltage difference of the power supply.
Replacing the Voltages with the capacitance equation gives
Q/C = Q1/C1 + Q2/C2 + Q3/C3
Because each plate of the capacitors is charged by either the power supply or by induction from the power supply, the value of Q1, Q2, Q3, and Q must all be the same. Simplifying gives
1/C = 1/C1 + 1/C2 + 1/C3. Given that C1, C2, and C3 are each equal to 8 microFarads this produces
1C = 3/8microF. Taking the recipracol gives
C = 8/3 = 2.67 microFarads for the total capacitance.