Ok - fair enough - so all of the information about the single point charge and its distance from the charged surface is moot (at least if we assume it is meant to be too small to matter for this level of course); the charge density is considered to be the charge over both surfaces (as stated)...
Thinking about it - I guess the 'charge density' must be the overall charge divided by the overall area - so 10/20 = 1/2. So E = (q/A) / (2e0) = 1/(4*e0) = 1/(4*8.854*10^-12) = 3.542*10^-11N/C.
Thanks again BvU. That's kind of what I thought - that this seems like it might be more complex that the extent of the course (AP Physics 1 - which is pre-cal). I didn't calculate σ based on the total charge (I think incorrectly using the point charge), so if I were to ignore the point charge...
Thanks for the feedback. Yes I realized that I missed the units on this one (only this one!) but even my one of the professors (who I finally managed to get in touch with and had tried to solve the problem) couldn't get the same answer as the answer key. So I think we must be right and the...
Thanks for your response BvU. Firstly, am I correct that the '2m away' is not relevant? If this is then the only equation to use, then I was marked wrong by the professor (they do not provide any feedback on wrong answers). The answer I gave was:
E = σ/2e0
=> E = 300/2*(8.854*10^-12)
=> E =...
I thought the equation listed should be used, with the 'charge density' determined by the point charge multiplied by the area of the plate, but not sure if that makes sense.
I thought it might be the case that the "2m away" wasn't applicable as the electric field doesn't change if the point away is less than the length of the plate, so I thought I should use the equation listed. All examples I can find talk about two charged plates, or the effect on cylinders...