Initially, we have two separate capacitors connected to voltage $V$:

• Capacitor 1 ($C$): Charge $Q_1 = CV$.
• Capacitor 2 ($2C$): Charge $Q_2 = 2CV$.
• Total Initial Charge: $Q_{initial} = 3CV$.
• Total Initial Energy: $U_{initial} = \frac{1}{2}(C+2C)V^2 = \frac{3}{2}CV^2$.

The smaller volume capacitor ($2C$) implies a smaller plate separation $d$ compared to the larger capacitor ($C$) with separation $D$. Since $C \propto 1/d$, $2C$ corresponds to separation $D/2$.

When inserted symmetrically, the setup forms three capacitance regions (gaps) as shown in Figure 3:

• Gap 1 (Top): Separation $D/4$. Capacitance becomes $4C$. Potential difference is $V$ (Outer $V$ to Inner $0$). Charge $q_1 = 4CV$.
• Gap 2 (Middle): Separation $D/2$. This is the inserted capacitor volume. Capacitance is $2C$. Potential difference is $V$ (Inner $0$ to Inner $V$). Charge $q_2 = 2CV$.
• Gap 3 (Bottom): Separation $D/4$. Capacitance becomes $4C$. Potential difference is $V$ (Inner $V$ to Outer $0$). Charge $q_3 = 4CV$.

The total charge supplied by the battery is the sum of charges on all plates connected to the positive terminal.

• Connected to Positive (+): Top Outer Plate ($q_1$) and Bottom Inner Plate ($q_2$ on top face, $q_3$ on bottom face).
• Charge on Top Outer Plate = $+4CV$.
• Charge on Bottom Inner Plate = $+2CV$ (facing up) + $+4CV$ (facing down) = $+6CV$.
• Total Final Charge: $Q_{final} = 4CV + 6CV = 10CV$.

Charge Flown through Battery: $$ \Delta Q = Q_{final} – Q_{initial} = 10CV – 3CV = 7CV $$

Since $\Delta Q$ is positive, charge flows out of the positive terminal. The battery does positive work on the circuit, meaning it delivers energy.