Post a New Question

physics 6a

posted by .

A potential difference of 4.75 kV is established between parallel plates in airf the air becomes ionized (and hence electrically conducting) when the electric field exceeds 3.00×106 V/m, what is the minimum separation the plates can have without ionizing the air?
d=___________m

  • physics 6a -

    V/d(min) = 3*10^6 V/m, the breakdown potential

    d(min) = (4.75*10^3 V)/(3*10^6 V/m)
    = 1.58*10^-3 m
    = 1.58 mm

Answer This Question

First Name
School Subject
Your Answer

Related Questions

More Related Questions

Post a New Question