Electric field between capacitors [closed]

A parallel-plate capacitor consists of two parallel, conducting plates of area $A$, separated by a distance $d$. Each carries a charge of magnitude $Q$; positive on one, negative on the other. Using Gauss' Law, find the electric field between the plates. (Indicate direction and magnitude.)

My solution: $E_1 A = \frac{Q}{\epsilon_0 \pi r^2}$; $E_2 A = \frac{-Q}{\epsilon_0 \pi r^2}$ (using a Gaussian cylinder)

Now, then $E = \frac{2Q}{\epsilon_0 \pi r^2}$.

I feel that I may have not taken some concepts into account. Namely, the distance d between the plates... I would definitely appreciate some suggestions.

closed as off-topic by John Rennie, Kyle Kanos, ACuriousMind♦, HDE 226868, tomJul 27 '15 at 21:15

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• wouldnt strictly call that field-theory more electrostatics. Anyway, your solution makes little sense (maybe you meant sth else) $|EA| = |Q|/\epsilon$ the area is the crossection, i.e. A (you draw a little rectangular box around one of the plates, not a cylinder, wrong symmetry). You do this only for one plate (how else with a closed surface, be careful re signs (surface normal parallel/antiparallel to field?)) . The distance does enter if you write it as a function of the voltage (not the charge) as field is nothing but voltage per distance: $E=V/d$. Both results allow to determine capacity – Bort Jul 27 '15 at 9:36
• Please note that Physics.StackExchange is not a homework help site. Please read this Meta post on asking homework-like questions and this Meta post for "check my work" problems. – Kyle Kanos Jul 27 '15 at 12:49

Indeed, the $\vec{E}$ field in a parallel plate is independent of distance from the plate. This works because of the assumption $d \ll$ length of plate (thus, we can ignore side effects of the plate). And as Bort pointed out, it is the Voltage $V$ that scales linearly with respect to distance from the plate, while $\vec{E}$ will remain constant.