Final answer:
The electric field between parallel conducting plates with a potential difference of 5.0 x 10^3 V and separated by 2.00 mm will be 2.5 x 10^6 V/m, which is less than the air's breakdown strength of 3.0 x 10^6 V/m.
Step-by-step explanation:
The question involves electric fields, potential difference, and the breakdown strength for air in the context of parallel conducting plates and high-voltage transmission lines. When considering the electric field between two plates with a potential difference of 5.0 x 103 V separated by 2.00 mm, the electric field strength (E) can be calculated using the formula E = V/d, where V is the potential difference and d is the distance between the plates. The electric field strength would be 2.5 x 106 V/m, which is below the breakdown strength for air (3.0 x 106 V/m), thus the electric field strength will not exceed the breakdown limit. For the high voltage transmission line, the potential difference between the plates and the potential at specific distances from the plate can be determined using the same concept of electric field and potential difference.