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 \( C_{\text{eq}} = \frac{C}{n-1} \).

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 \(5\text{ N}\).

In a charged capacitor, the electrostatic potential energy is stored in the electric field that exists between the plates of the capacitor. The energy density is given by \(u_E = \frac{1}{2}\epsilon_0 E^2\).

The energy stored in the capacitor is \(U = \frac{1}{2} C V^2 = \frac{1}{2} \times 40 \times 10^{-6} \times (3000)^2 = 180\text{ J}\). Power delivered is \(P = \frac{U}{t} = \frac{180}{2 \times 10^{-3}} = 90\text{ kW}\).

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 \(W_b = \Delta Q \cdot V = CV^2\).

Dielectric strength is the maximum field \(E_{\text{max}} = 2\text{ kV/mm} = 2 \times 10^6\text{ V/m}\). For thickness \(d = 50 \mu\text{m} = 5 \times 10^{-5}\text{ m}\), the maximum potential difference is \(V_{\text{max}} = E_{\text{max}} \times d = 2 \times 10^6 \times 5 \times 10^{-5} = 100\text{ V}\).

The energy stored is \(E = \frac{1}{2} C V^2\). With the battery connected, potential \(V\) is constant. Doubling the separation halves the capacitance, so the new capacitance is \(C' = C/2\), and the stored energy becomes \(E' = E/2 = 0.50 E\).

When a capacitor is disconnected, charge (Q) remains constant. Capacitance \(C = \frac{\epsilon_0 A}{d}\). If (d) doubles, (C) halves. Stored energy \(U = \frac{Q^2}{2C}\). Since (Q) is constant and (C) halves, (U) doubles.