Question

Find the maximum potential difference between two parallel conducting plates separated by 0.500 cm of air, given the maximum sustainable electric field strength in air to be (3.0 times 10^6 V/m).

A) (1.5 kV)
B) (3.0 kV)
C) (4.5 kV)
D) (6.0 kV)

Answer 1

The maximum potential difference between two parallel conducting plates separated by 0.500 cm of air is 1.5 kV.

Step-by-step explanation:

The maximum potential difference between two parallel conducting plates separated by 0.500 cm of air can be found using the equation V = Ed, where V represents the potential difference, E represents the electric field strength, and d represents the distance between the plates. In this case, the maximum sustainable electric field strength in air is given as 3.0 × 106 V/m. Thus, we can substitute this value into the equation along with the distance of 0.500 cm (which can be converted to meters) to find the maximum potential difference. Multiplying the electric field strength and the distance will give us a maximum potential difference of 1.5 kV. Therefore, the correct answer is A) (1.5 kV).