Find the potential difference V_a - V_b between nodes a and b for...

In the circuit labeled (c), let the potential at the bottom middle node $b$ be $V_b = 0$.

The left battery has an electromotive force of $2 \text{ V}$ with its positive terminal facing right. The potential at the node between the left battery and the left $2 \mu\text{F}$ capacitor is $V_a - 2$.

The right battery also has an electromotive force of $2 \text{ V}$ with its positive terminal facing right (indicated by the longer line). The potential at the node between the right battery and the right $2 \mu\text{F}$ capacitor is $V_a + 2$.

Let the capacitance of the middle capacitor be $C$. Applying charge conservation for the isolated plates connected to node $b$:

$2(V_b - (V_a - 2)) + 2(V_b - (V_a + 2)) + C(V_b - V_a) = 0$

Substituting $V_b = 0$ yields $2(-V_a + 2) + 2(-V_a - 2) + C(-V_a) = 0$.

$-2V_a + 4 - 2V_a - 4 - CV_a = 0$

$-(4 + C)V_a = 0 \implies V_a = 0$.

Therefore, the potential difference is $V_a - V_b = 0 - 0 = 0$.