A 5.0 µF capacitor rated 150 V is joined in series with a 20 µF capacitor rated 400 V…

Step 1. The equivalent. $C = \dfrac{5.0\ \mu\text{F} \times 20\ \mu\text{F}}{5.0\ \mu\text{F} + 20\ \mu\text{F}} = 4.0\ \mu\text{F}$.

Step 2. The charge limits. The $5.0\ \mu\text{F}$ capacitor reaches $150\ \text{V}$ at $5.0\ \mu\text{F} \times 150\ \text{V} = 750\ \mu\text{C}$. The $20\ \mu\text{F}$ capacitor reaches $400\ \text{V}$ at $20\ \mu\text{F} \times 400\ \text{V} = 8000\ \mu\text{C}$.

Step 3. The smaller one wins. The pair can hold $750\ \mu\text{C}$ at most.

Step 4. The voltages. The $5.0\ \mu\text{F}$ capacitor has $150\ \text{V}$. The $20\ \mu\text{F}$ capacitor has $\dfrac{750\ \mu\text{C}}{20\ \mu\text{F}} = 37.5\ \text{V}$. Together, $150\ \text{V} + 37.5\ \text{V} = 187.5\ \text{V}$.