Introduction:
A parallel plate capacitor consists of two conducting plates separated by a distance. The maximum potential difference that can be applied to the capacitor depends on several factors, including the distance between the plates and the dielectric material between them.

Given information:
- Distance between the plates (d) = 0.1 mm
- Medium between the plates is air

Determining the maximum potential difference:
The maximum potential difference that can be applied to a parallel plate capacitor without causing dielectric breakdown or discharge is determined by the breakdown voltage or dielectric strength of the medium between the plates.

Breakdown voltage of air:
The breakdown voltage of air is approximately 3 million volts per meter (3 MV/m) under normal conditions. This means that if the electric field strength exceeds 3 MV/m, the air between the plates will break down and allow current to flow, causing a discharge.

Calculating the maximum potential difference:
To calculate the maximum potential difference that can be applied to the capacitor, we need to convert the distance between the plates from millimeters to meters.

Given: d = 0.1 mm

Converting to meters: d = 0.1 mm = 0.1 x 10^-3 m = 10^-4 m

Applying the formula:
The electric field strength (E) between the plates of a parallel plate capacitor is given by the formula:

E = V / d

Where:
- E is the electric field strength
- V is the potential difference between the plates
- d is the distance between the plates

Rearranging the formula to solve for V:

V = E x d

Using the breakdown voltage of air (3 MV/m) as the maximum electric field strength:

V = 3 x 10^6 V/m x 10^-4 m

V = 3 x 10^2 V

Final answer:
The maximum potential difference that can be applied to the parallel plate capacitor is 300 volts.