A

Step 1: Understand the concept of pressure in a dielectric.
In a parallel-plate capacitor filled with a dielectric (in this case, water), the electric field ($E$) between the plates will contribute to the pressure change in the dielectric. The electric field can be calculated using the equation:
$$E = \frac{V}{d}$$
where
- $V$ = Voltage applied (500 V)
- $d$ = distance between the plates (1.0 mm = 0.001 m)

Step 2: Calculate the electric field.
$$E = \frac{500 \, \text{V}}{0.001 \, \text{m}} = 500000 \, \text{V/m}$$

Step 3: Relate electric field to pressure increment.
The pressure increment ($\Delta P$) in a dielectric is given by the equation:
$$\Delta P = \frac{1}{2} \epsilon_0 E^2$$
where:
- $\epsilon_0$ = permittivity of free space ($8.85 \times 10^{-12} \, F/m$)

Step 4: Substitute the values into the equation.
- First, calculate $E^2$:
$$E^2 = (500000 \, \text{V/m})^2 = 250000000000000 \, \text{V}^2/m^2$$
- Then, calculate $\Delta P$:
$$\Delta P = \frac{1}{2} \times (8.85 \times 10^{-12} \, F/m) \times (250000000000000 \, \text{V}^2/m^2)$$
$$\Delta P = \frac{1}{2} \times 8.85 \times 10^{-12} \times 250000000000000$$
$$\Delta P \approx 1.10625 \times 10^{3} \, \text{Pa}$$

Step 5: Conclusion.
Therefore, the pressure increment in the gap is approximately 1106.25 Pa.
Hence, the correct answer is A.