obvious question IMO, but just need to make sure before test. if you have two parallel conducting plates, and apply charge to one of them, then the field between them is half, compared to if the plates were acting as an air capacitor ie. both plates oppositely charged? thing is, when I use a simplified version of gauss, the field on an infinite sheet of charge is charge density/permittivity which is the exact same as if I use Q/V=[permittivity*plate area]/plate seperation. i tried to use latex, but i've never really used it, and it came out all wrong due bad syntax. sorry.