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, \(Phi_E = 0\).

Initially, the uncharged ball is attracted to the positive plate by induction. Upon touching, it acquires positive charge and is repelled by the positive plate, moving towards the negative plate. Upon touching the negative plate, it acquires negative charge and is repelled by the negative plate, moving back to the positive plate. This continuous charge transfer and repulsion causes the ball to oscillate between the plates, touching each in turn.