Consider two capacitors, each possessing a capacitance C and a...

For two capacitors connected in series, the equivalent capacitance $C_{eq}$ is given by:

$\dfrac{1}{C_{eq}} = \dfrac{1}{C_1} + \dfrac{1}{C_2}$

Since $C_1 = C_2 = C$, we have:

$\dfrac{1}{C_{eq}} = \dfrac{1}{C} + \dfrac{1}{C} = \dfrac{2}{C}$

$C_{eq} = \dfrac{C}{2}$

In a series combination of identical capacitors, the total applied voltage is divided equally among them. If $V_{total}$ is the voltage applied across the combination, the voltage across each capacitor is $\dfrac{V_{total}}{2}$.

For the capacitors not to break down, the voltage across each must not exceed its individual breakdown voltage $V$.

$\dfrac{V_{total}}{2} \le V \Rightarrow V_{total} \le 2V$

Thus, the breakdown voltage of the series combination is $2V$.