**Introduction:**
In this scenario, we have two capacitors, A and B. Capacitor A is filled with a dielectric whose dielectric constant varies with the applied voltage, K = V. Capacitor B is identical to A, but it has air as the dielectric. The capacitors are connected to a voltage source and then connected to each other.

**Explanation:**

**1. Capacitor A**
- Capacitor A is filled with a dielectric whose dielectric constant varies with the applied voltage, K = V.
- This means that as the voltage across capacitor A increases, its dielectric constant also increases.
- The capacitance of capacitor A can be calculated using the formula: C = K * ε₀ * A / d, where C is the capacitance, K is the dielectric constant, ε₀ is the permittivity of free space, A is the area of the plates, and d is the distance between the plates.
- Since K = V, the capacitance of capacitor A can be expressed as: C = V * ε₀ * A / d.

**2. Capacitor B**
- Capacitor B is identical to A, but it has air as the dielectric.
- The capacitance of capacitor B can be calculated using the same formula as capacitor A, but with K = 1 (dielectric constant of air): C = ε₀ * A / d.
- The capacitance of capacitor B is constant and does not depend on the applied voltage.

**3. Connection to Voltage Source**
- When capacitors A and B are connected to a voltage source, they initially store charge and build up a potential difference.
- The charge stored in a capacitor can be calculated using the formula: Q = C * V, where Q is the charge, C is the capacitance, and V is the voltage.
- As the voltage source is connected, both capacitors A and B will accumulate charge. The charge on capacitor A will be greater than that on capacitor B because capacitor A has a larger capacitance due to its variable dielectric constant.

**4. Connection of Capacitors A and B**
- After disconnecting the voltage source, capacitors A and B are connected to each other.
- The total charge on the system remains conserved. Therefore, the charge on capacitor A will distribute itself between the two capacitors.
- Since the capacitance of capacitor B is constant, the voltage across capacitor B remains the same.
- The voltage across capacitor A will decrease as it is connected to capacitor B.
- The final voltage across capacitor A and B depends on their initial charge and capacitance values.

In summary, capacitor A, filled with a dielectric whose dielectric constant varies with applied voltage, will have a larger capacitance and charge compared to capacitor B, which has air as the dielectric. When connected to each other after disconnecting the voltage source, the charge will distribute itself between the capacitors, and the voltage across capacitor A will decrease while the voltage across capacitor B remains the same.