Evaluate the potential difference V_a - V_b between points a and b...

In the circuit labeled (d), let the potential at node $a$ be $V_a = 0$.

The circuit consists of three parallel branches connected between nodes $a$ and $b$. Each branch contains a battery and a capacitor in series.

For all three batteries, the positive terminal (longer line) is oriented towards the right.

The potentials at the intermediate nodes between the batteries and capacitors are $6 \text{ V}$ (top branch), $12 \text{ V}$ (middle branch), and $24 \text{ V}$ (bottom branch).

Applying the principle of charge conservation for the isolated plates connected to node $b$:

$4(V_b - 6) + 2(V_b - 12) + 1(V_b - 24) = 0$

Expanding the equation gives $4V_b - 24 + 2V_b - 24 + V_b - 24 = 0$.

$7V_b - 72 = 0 \implies V_b = \dfrac{72}{7} \approx 10.28 \text{ V}$.

The potential difference is $V_a - V_b = 0 - 10.28 = -10.28 \text{ V}$, which corresponds to $-10.3 \text{ V}$.