The potential difference depends on the charge on the inner facing surfaces, which is given by \( q = \frac{q_1 - q_2}{2} = \frac{2.0 \times 10^{-8} - (-1.0 \times 10^{-8})}{2} = 1.5 \times 10^{-8}\text{ C} \). Using \( V = \frac{q}{C} \), we get \( V = \frac{1.5 \times 10^{-8}}{1.2 \times 10^{-9}} = 12.5\text{ V} \).

If the plates are connected in series, the total capacitance \( C_{\text{eq}} \) of \( n-1 \) capacitors is given by \( \frac{1}{C_{\text{eq}}} = \frac{n-1}{C} \), which yields \( C_{\text{eq}} = \frac{C}{n-1} \).

For the common potential to be zero, the initial charges on the capacitors must be equal in magnitude and connected with opposite polarities: \( q_1 = q_2 \Rightarrow 120 C_1 = 200 C_2 \Rightarrow 3 C_1 = 5 C_2 \).

The total charge is \( Q = 150 \mu\text{C} + 150 \mu\text{C} = 300 \mu\text{C} \). The total capacitance is \( C = 4piepsilon_0(R_1 + R_2) = \frac{0.3}{9 \times 10^9}\text{ F} \). The common potential is \( V = \frac{Q}{C} = \frac{300 \times 10^{-6} \times 9 \times 10^9}{0.3} = 9 \times 10^6\text{ V} \).

Since both capacitors are in series and have equal capacitance, the total potential difference divides equally between them. The maximum potential is limited by the weaker capacitor: \( V_{\text{max}} = 2 \times 4\text{V} = 8\text{V} \).

Since the capacitor is isolated, its charge \(Q\) remains constant. When the separation \(d\) increases, the capacitance \(C = \frac{\epsilon_0 A}{d}\) decreases. Since \(U = \frac{Q^2}{2C}\), the stored energy increases due to the work done against the electrostatic attraction.

The force of attraction between the plates of a parallel plate capacitor is given by \(F = \frac{Q^2}{2 \epsilon_0 A}\). Substituting the values: \(F = \frac{(5 \times 10^{-6})^2}{2 \times 8.85 \times 10^{-12} \times 0.3} \approx 4.7\text{ N}\), which is about \(5\text{ N}\).

When the battery is disconnected, the charge \(Q\) remains constant. The initial energy is \(U_i = \frac{Q^2}{2C_0}\). After inserting the dielectric of constant \(K\), the capacitance becomes \(C = K C_0\) and the final energy is \(U_f = \frac{Q^2}{2K C_0} = \frac{U_i}{K}\). Therefore, the ratio \(U_i / U_f = K\).

When the plate separation is halved, capacitance becomes \(2C\). Keeping charge constant at \(Q = CV\), to recharge it back to potential \(V\), the final charge is \(Q' = 2CV\). The charge flowing from the battery is \(\Delta Q = 2CV - CV = CV\). The energy supplied by the battery is \(W_b = \Delta Q \cdot V = CV^2\).

As the capacitor remains connected to the battery, the potential difference \(V\) remains constant. The stored energy is \(U = \frac{1}{2} C V^2\). Since slipping a dielectric slab increases the capacitance \(C\), the energy increases correspondingly.