Two parallel plates, connected to a 45-V power supply, are separated by an air gap. How small can the gap be if the air is not to become conducting by exceeding its breakdown value of $E = 3 \times 10^6 \textrm{ V/m}$?

VIDEO TRANSCRIPT

This is Giancoli Answers with Mr. Dychko. Air of all have a dielectric breakdown which means it will ionize and start conducting. If the electric field through it is about 3 times 10 to the 6 volts per meter, and we can solve for the distance separating the final and the initial voltages by rearranging this formula. Electric field is the volts potential difference divided by the distance between the two points. And solving for d, so multiply both sides by d over e. Separation is potential difference divided by electric field. We want a potential difference of 45 volts and divided by three times 10 to the six volts per meter electric field, when you do have dielectric breakdown and the air starts conducting, this will work out to 1.5 times 10 the minus five meters or rounded to one significant figure, that would be about 20 micrometers. If the separation is 20 micrometers or less, then the air will be conducting. But if you separate it by more than 20 micrometers, then the air will not be conducting.