Final answer:
The electric field strength between two plates separated by 2.00 mm and with a 5,000 V potential difference is 2.5 × 10⁶ V/m, which does not exceed the breakdown strength for air. The plates can be as close as 1.67 mm with this voltage without exceeding the breakdown limit.
Step-by-step explanation:
The question involves calculating the electric field strength between two parallel conducting plates and determining if it will exceed the breakdown strength for air.
(a) To check if the electric field strength will exceed the breakdown strength for air when a potential difference of 5,000 V is applied and the plates are separated by 2.00 mm, we use the electric field equation E = V/d. Here, V = 5,000 V and d = 2.00 mm = 0.002 meters:
E = 5,000 V / 0.002 m = 2.5 × 10⁶ V/m.
Since the calculated electric field strength of 2.5 × 10⁶ V/m is less than the breakdown strength for air, which is 3.0 × 10⁶ V/m, the electric field does not exceed the breakdown limit.
(b) To find out how close together the plates can be with the applied voltage of 5,000 V without exceeding the breakdown limit:
The maximum permissible electric field strength is 3.0 × 10⁶ V/m. Using the electric field equation E = V/d, and rearranging for d gives d = V/E:
d = 5,000 V / (3.0 × 10⁶ V/m) = 0.00167 m or 1.67 mm.
Therefore, with a potential difference of 5,000 V, the plates can be as close together as 1.67 mm without exceeding the breakdown strength for air.