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Parallel insulating plates

  1. Jan 26, 2010 #1
    Electric Fields/Gauss's Law: 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?
     
    Last edited: Jan 26, 2010
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  3. Jan 26, 2010 #2

    Gokul43201

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    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?
     
  4. Jan 26, 2010 #3
    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
     
    Last edited: Jan 26, 2010
  5. Jan 26, 2010 #4

    Gokul43201

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    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?
     
  6. Jan 26, 2010 #5
    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.
     
  7. Jan 27, 2010 #6

    Gokul43201

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    Coulomb's Law does apply, so long as you apply it correctly. What is important about Coulomb's Law is that it applies only to a single point charge. To calculate electric fields from a distribution of charges via Coulombs law, you would have to integrate the electric field (which is a vector quantity, so you have to be careful with this) due to each of the charges spread out over the entire surface containing them. This can be a messy thing to do, which is why it is often easier to utilize the symmetry of the problem and apply Gauss' Law.

    Furthermore, the Gaussian surface, is not a plane. Gauss' Law tells you about the flux through a closed surface, and a plane is not a closed surface. Look up "Gaussian pillbox".
     
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