## Parallel insulating plates

Here's the question: Two large, parallel, insulating plates are charged uniformly with the same charge density σ. What is the magnitude of the resultant electric field E ?
The correct answer: zero between the plates, σ/E0 outside
My question: Why is the field between them zero? I understand that there is at least a point in between them where the field is zero, but why is the whole field zero. As you get closer to one plate the force from that plate increases (whether it be attractive or repulsive) and the force from the other decreases. This means that the electric field is changing (the only way to get an unbalanced force). In addition, if the two add together, why do you get σ/E0 for the answer. Shouldn't each have a field vector of σ/E0 and when they add, 2σ/E0? I don't really see how they get σ/2E0 for each plate when they each have a charge density of σ. Finally, why does the field not vary with the distance? Is it just a property of a plate or am I just missing something?
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 Recognitions: Gold Member Science Advisor Staff Emeritus Yes, you are missing something. You are missing the approach you need to take to solve the problem. You can not just guess what the field is going to do with distance, you should calculate it using concepts you have learned. What useful concept/theorem could you use to calculate the electric field from a nice, symmetric distribution of charge, such as in this problem?

 Quote by Gokul43201 What useful concept/theorem could you use to calculate the electric field from a nice, symmetric distribution of charge, such as in this problem?
I'm guessing you are talking about Gauss's Law, but if you integrate, you get the integral of 2ko<pi>dr/r2. Therefore the force does vary by difference, that's Coulomb's Law

I'm not saying you're wrong, but I'm just finding things conflicting with each other

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## Parallel insulating plates

 Quote by LonghornDude8 I'm guessing you are talking about Gauss's Law, but if you integrate, you get the integral of 2kodr/r2. Therefore the force does vary by difference, that's Coulomb's Law
No, that's not correct. Do this carefully, and one step at a time. If you use Gauss' Law correctly, you end up having to do little or no integration. And in any case, please be as descriptive as possible in explaining yourself.

First, when applying Gauss' Law, you need to begin by constructing a Gaussian surface. What is the shape of the surface, and where is it constructed? Then what's the next step?

 Quote by Gokul43201 First, when applying Gauss' Law, you need to begin by constructing a Gaussian surface. What is the shape of the surface, and where is it constructed? Then what's the next step?
The Gaussian surface is a plane (flat plane) and I actually just found the powerpoint from my teacher and there was a slide on this. I understand what you need to do to solve the problem, I just don't really understand why Coulomb's law does not apply.

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