Question

A potential difference of 4.50 kV is established between parallel plates in air.

If the air becomes ionized (and hence electrically conducting) when the electric field exceeds 3.05×10⁶ V/m , what is the minimum separation the plates can have without ionizing the air?

Answer in mm.

Answer 1

Potential difference (ΔV) between the plates of the capacitor = 4.50 kV

                                                                                                 = 4.50 x 10³ V

Electric field (E) between the plates of the capacitor = 3.05 x 10⁶ V/m

Then the minimum separation (d) between the plates of the capacitor is calculated as

                  E = ΔV / d

                  d = ΔV / E

                     = (4.50 x 10³ V) / (3.05 x 10⁶ V/m)

                     = 1.475 x 10^-3 m

                     = 1.475 mm