Three capacitors each of capacitance $C$ and of breakdown voltage $V$ are joined in series. The capacitance and breakdown voltage of the combination will be:

1. Capacitance in series:
$$\frac{1}{C_{eq}}=\frac{1}{C}+\frac{1}{C}+\frac{1}{C}=\frac{3}{C}⟹C_{eq}=\frac{C}{3}$$
2. Breakdown voltage in series:In series, the total voltage applied is shared across the capacitors. If each can withstand a maximum of $V$, the total combination can withstand:
$$V_{total}=V+V+V=3V$$