A potential difference of 3.75 kV is established between parallel plates in air What is the surface charge density on each plate?

Answer 1

"Without the size of the plates (and a little additional information), we're dead in the water on this. Charge density is charge (Q) divided by the area of the capacitor's plates (a)."

With some critical thinking, you can figure out the charge density without the size of the plates. I'm assuming you had the same physics problem I did, in which another part makes you assume you're at minimum separation, in which case you can use the formula for electric field between two equally and oppositely charged plates (E = sigma/epsilon_0, where sigma is charge density and epsilon_0 is permittivity of free space, 8.85*10^-12 F/m). Using this knowledge, as well as the idea that E = potential difference / separation (E = V/d). So you rearrange the first equation to solve for sigma and then use E = V/d with the values the problem gives you to solve (sigma = epsilon_0*V/d). Your question did not give a distance, so I hope the problem did...haha... Or it could have been like mine and made you solve for the minimum separation first, using the potential and the electric field that results in air becoming electrically conductive. In that case, you take the former value divided by the latter and you get the minimum separation and use that value as your distance.