1. Step 1: Charging

Source $V_0$ connected to A and D. B and C connected together.
This creates two capacitors in series: $C_{AB}$ and $C_{CD}$ (since $V_B=V_C$, the gap B-C has no field).
Charge $Q_0 = C_{eq} V_0 = \frac{C}{2} V_0$.
Charge distribution: Plate A ($+Q_0$), Plate D ($-Q_0$). Plates B and C acquire induced charges $-Q_0$ and $+Q_0$ on their outer faces respectively.

2. Step 2 & 3: Isolation

Wire B-C removed: Charges on B and C are trapped. $Q_B = -Q_0$, $Q_C = +Q_0$.
Source removed: Charges on A and D are trapped (open circuit). $Q_A = +Q_0$, $Q_D = -Q_0$.
Here $Q_0 = \frac{\epsilon_0 A V_0}{2d}$.

3. Step 4: Connecting A and D

Wire connects A and D: $V_A = V_D$. Charge redistributes between A and D, but total charge $Q_A+Q_D = 0$ is conserved.
Let charge $x$ be on the inner face of A.
Distribution logic implies electric fields in gaps are proportional to:
Gap AB: $x$
Gap BC: $-Q_0 + x$
Gap CD: $x$
Condition $\sum \Delta V = 0 \implies x + (-Q_0 + x) + x = 0 \implies 3x = Q_0 \implies x = Q_0/3$.

4. Calculating Potentials

Potential drop $\Delta V = \frac{Q_{face} d}{A \epsilon_0}$. Recall $V_0 = \frac{2 Q_0 d}{A \epsilon_0}$.

• $V_A – V_B = \frac{(Q_0/3) d}{A \epsilon_0} = \frac{1}{3} \frac{V_0}{2} = \frac{V_0}{6}$.
• $V_B – V_C = \frac{(-2Q_0/3) d}{A \epsilon_0} = -\frac{2}{3} \frac{V_0}{2} = -\frac{V_0}{3}$.
• $V_C – V_D = \frac{V_0}{6}$.