Question
A potential difference of 3.75KV is established between parallel plates in air. If the air becomes electrically conducting when the electric field exceeds 3x106 V/m. What is the minimum separation of the plates? When the separation has the minimum value, what is the surface charge density on each plate?
Answers
Anonymous
3*10^6 * x = 37,500
solve for x
s = charge density = eo * E
= 8.85 *10^-12Farads/meter * 3*10^6 volts/meter
solve for x
s = charge density = eo * E
= 8.85 *10^-12Farads/meter * 3*10^6 volts/meter