Final answer:
The total capacitance of the capacitors connected in series is 0.325 µF.
Step-by-step explanation:
The total capacitance of capacitors connected in series can be found by using the equation for capacitance in series:
cs = 1/((1/1.000) + (1/5.000) + (1/8.000))
Substituting the given capacitances into the equation, we get:
cs = 1/((1/1.000) + (1/5.000) + (1/8.000)) = 0.325 µF
Therefore, the total capacitance of the three capacitors connected in series is 0.325 µF.