Two capacitors joined in series and connected to a battery are...

For capacitors connected in series, the charge $Q$ on each capacitor is the same.

The potential difference $V$ across a capacitor is given by $V = \dfrac{Q}{C}$. Since $Q$ is constant for both capacitors, the potential difference is inversely proportional to the capacitance ($V \propto \dfrac{1}{C}$).

From the given $V-x$ graph, the vertical rise represents the potential difference across each capacitor. The graph clearly shows that the potential difference across the first capacitor ($V_1$) is greater than the potential difference across the second capacitor ($V_2$).

Therefore, $V_1 > V_2$.

Substituting $V = \dfrac{Q}{C}$, we get:

$\dfrac{Q}{C_1} > \dfrac{Q}{C_2}$

Since $Q$ is positive, this inequality simplifies to:

$C_1 < C_2$