Final answer:
The maximum electric field strength dry air can support is approximately 3.0×106 V/m. For two parallel conducting plates, if separated by 0.500 cm, the maximum potential difference before ionization occurs is 15,000 V. The calculation involves multiplying the electric field strength by the separation distance between the plates.
Step-by-step explanation:
The question pertains to the calculation of the maximum potential difference between two parallel conducting plates in an electric field. According to the problem given, we know that the maximum electric field strength that dry air can sustain is about 3.0×106 V/m before ionization occurs. The torque on a sphere will not be directly affected by the electric field between the plates since torque is related to forces acting on the sphere rather than the potential difference across the plates. However, if we wanted to calculate the maximum voltage (potential difference) that can be applied between two plates without causing a discharge through the air, we'd use the equation V = E×d, where V is the voltage, E is the electric field strength, and d is the separation between the plates.
For example, if the plates are separated by 0.500 cm, or 0.005 meters, the maximum voltage that can be applied across them without causing ionization of the air is calculated as follows:
- Electric field strength, E = 3.0×106 V/m
- Distance, d = 0.005 m
- Maximum Voltage, V = E×d = 3.0×106 V/m × 0.005 m = 15,000 V
This calculation shows that the maximum voltage that can be sustained between two parallel plates separated by 0.500 cm in air is 15,000 V, which is the maximum potential difference before air becomes conductive.