# 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

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – John Rennie, Kyle Kanos, ACuriousMind, HDE 226868, tom
If this question can be reworded to fit the rules in the help center, please edit the question.

• 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.