The solution can be derived as follows:

1. Initial Capacitance (with air as the dielectric):
   For a parallel plate capacitor with air as the dielectric, the capacitance $C_{\text{initial}}$ 

   is given by: 

   $$C_{\text{initial}} = C$$The battery applies a potential difference

   $E$, so the initial charge on the capacitor is:

    

   $$Q_{\text{initial}} = C_{\text{initial}} \times E = C \times E$$ 

2. Capacitance with Dielectric Slab:
   When a dielectric slab of dielectric constant $K$ 

   is inserted between the plates, the capacitance increases by a factor of $K$ 

   , so the new capacitance $C_{\text{new}}$ 

   becomes: 

   $$C_{\text{new}} = K \times C$$ 

3. Final Charge on Capacitor:
   The battery remains connected and maintains a constant voltage $E$ 

   . Therefore, the final charge on the capacitor is: 

   $$Q_{\text{final}} = C_{\text{new}} \times E = K \times C \times E$$ 

4. Additional Charge Drawn from the Battery:
   The additional charge drawn from the battery is the difference between the final charge and the initial charge: 

   $$\Delta Q = Q_{\text{final}} - Q_{\text{initial}} = (K \times C \times E) - (C \times E)$$Simplifying:

    

   $$\Delta Q = (K - 1) \times C \times E$$ 

Thus, the additional charge drawn from the battery is:

$$(K - 1) \times C \times E$$

Energy loss when connecting two capacitors is \( \Delta U = \frac{1}{2} \frac{C_1 C_2}{C_1 + C_2} (V_1 - V_2)^2 \). Here, \( C_1 = C_2 = 400\text{ pF} \), \( V_1 = 100\text{ V} \), and \( V_2 = 0 \). This gives \( \Delta U = \frac{1}{4} C V^2 = 1 \times 10^{-6}\text{ J} \).

Energy density (energy per unit volume) of a capacitor is given by the formula \( u = \frac{1}{2} \varepsilon_0 E^2 \). Substituting electric field \( E = \frac{V}{d} \) gives \( u = \frac{1}{2} \varepsilon_0 \frac{V^2}{d^2} \).

The capacitance of a capacitor is proportional to the permittivity of the medium. Thus, when air is replaced by a dielectric of constant K, capacity increases K times.

For an isolated capacitor, force is \(F = \frac{Q^2}{2 K A \varepsilon_0}\) which decreases with dielectric (\(K > 1\)). The electric field \(E = \frac{E_0}{K}\) also decreases. Hence, Reason is false.

Let \(r\) be the radius of a small drop and \(R\) be the radius of the big drop. Volume conservation: \(8 \times \frac{4}{3}\pi r^3 = \frac{4}{3}\pi R^3 ⇒ R = 2r\). Capacitance of a spherical drop is \(C = 4\pi\epsilon_0 R\). So, \(C_{\text{big}} = 4\pi\epsilon_0 (2r) = 2 (4\pi\epsilon_0 r) = 2 C_{\text{small}}\).

For an isolated capacitor, the charge \(Q\) remains constant. \(Q = C_1 V_1 = C_2 V_2\). Since \(C = \frac{\epsilon_0 A}{d}\), we have \( \frac{\epsilon_0 A}{d_1} V_1 = \frac{\epsilon_0 A}{d_2} V_2 \implies \frac{V_1}{d_1} = \frac{V_2}{d_2} \). Thus, \(V_2 = V_1 \frac{d_2}{d_1} = 60 \text{ V} \times \frac{12 \text{ mm}}{4 \text{ mm}} = 180 \text{ V}\).

When a total charge \(Q_{\text{total}} = 2 \mu C\) is given to one plate, the effective charge stored on the capacitor plates is \(Q = \frac{Q_{\text{total}}}{2} = \frac{2 \mu C}{2} = 1 \mu C\). Given capacitance \(C = 1 \mu F\). Potential difference \(V = \frac{Q}{C} = \frac{1 \mu C}{1 \mu F} = 1 \text{ V}\).

A capacitor consists of two plates with equal and opposite charges, \(+Q\) and \(-Q\), where \(Q = CV\). If a closed surface encloses the entire capacitor, the net charge enclosed within the surface is \(Q_{\text{enclosed}} = (+Q) + (-Q) = 0\). By Gauss's Law, the total electric flux through this closed surface is \(Phi_E = \frac{Q_{\text{enclosed}}}{\epsilon_0}\). Therefore, \(Phi_E = 0\).

The capacitance of a capacitor is a constant determined by its geometric shape, size, and the medium between the plates. It is independent of the charge \( Q \) and potential difference \( V \) applied to it.