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Why is Electric Field Constant between a Parallel Plate Capacitor?

  1. Nov 13, 2014 #1
    So electric field tells us the force per unit charge that is felt by a test charge at a distance d from a source charge.
    So it tells us that the closer the test, or other charge, is to the source charge ,the stronger the interaction, and also that the larger the source charge, the stronger the interaction as well.

    Hence E = F/q = kQsource/d^2

    1) What I am having trouble understanding is how the electric field between a parallel plate capacitor is constant?

    I think part of my misunderstanding comes from the fact that we're not dealing with a single source charge but an electric field created by the interaction between two sources, the positive plate and the negative plate. Is that interaction what makes the electric field constant?

    2) Also, while electric field changes with distance from a source charge, in between a parallel plate capacitor, the electric field is constant regardless of where you are in between the capacitor?

    3) To add to that, the electric potential decreases as you go from positive to negative plate, yet the electric field doesnt. Normally, a change in electric potential changes the electric field doesnt it?
     
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  3. Nov 13, 2014 #2

    Matterwave

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    1) No, the electric field from a single infinite plate is constant as well. That the electric field inside a plate capacitor is constant is only an approximation. This works for distances very close to the plates, and when you are far away from the edges of the plates. In other words, it works for plates which are separated by a distance, d, which is much smaller than the typical size of the plate (e.g. d<<r where r is the radius of a plate, for a circular capacitor plate, or d<<w,l for a square capacitor plate where w and l are the width and length of the plate, respectively). In the case that we are very close to the plate, we can treat the plate as infinite, and the result from the infinite plate takes over.

    2) As long as the plate separation is small, and you are away from the edges of the plates. This is a result of the geometry of the source charges. It is not a single source charge, it is a whole infinite plane of source charges. To see this, one can use Gauss's law.

    3) No, you do not need the electric field to change for the electric potential to change. The electric potential difference between two points is basically the line integral of the electric field from one point to the next, and the integral of a constant function will still change if you change the two end points.
     
  4. Nov 13, 2014 #3
    What is the significance of the distance between the plate and being able to make the approximation?

    For a single plate, why would the electric field be constant? Because normally, for point charges, the electric field lines are further apart the further yuo get away from the charge, and/or as you get further away, via the inverse square law, the electric field strength decreases with distance. Why is it different for a plate?
     
  5. Nov 13, 2014 #4

    Matterwave

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    Like I said, the plate has to be infinite for this to work. In order to be able to treat the plate as approximately infinite, you have to be very close to the plate, and very far from the edges of the plate.
     
  6. Nov 13, 2014 #5
    Imagine a point charge P very close to the plate. There will be a charge on the plate right under it which will pull it towards the plate. There are other charges farther away pulling it but at some angle. For charges far away from P, the force will be mostly horizontal and their effect will cancel (they are all around P). If you move P a little further from the plate, the effect of the charge right under it will decrease but the forces from far away charges will have larger normal components, even though the magnitudes of the forces decreases. Altogether these effects compensate to keep the field constant.
    It's like P is "seeing" (actualy interacting with ) more and more charge as it moves away from the plate, which balances the increased distance.
     
  7. Nov 14, 2014 #6
    imagine one of these old gymnastic instruments with two handles and stud (Gestüt) between them (like springs). Imagine these springs like the EF lines, All they have the same thickness. When you open it, the springs's thickness becomes smaller.
    The electric field is constand, means that the differences between same distances between 2 Punkte(i mean points) are the same. But the differences between difference distances between 2 Points are different.
     
  8. Nov 14, 2014 #7
    The statement that a single infinite plate has constant electric field can be argued through symmetry. Basically, for an ideally continuous, uniform and infinite plate, any configurations after arbitrary displacements are identical to that before. i.e. you cannot tell the difference and so cannot the field.

    In fact, a mathematically rigorous approach is fairly straight forward. For simplicity, just take a uniform disc and the field on axis could be found by integrating over concentric rings. And the result would show that the ratio d/R determines the deviation from uniform field (look up on web if you do not know how to do this). In the limit of infinite plate, it is actually uniform on "axis".
     
  9. Nov 14, 2014 #8
    The electric field between two capacitor plates is approximately constant. You need to have an understanding of when the approximation is valid.

    I think the easiest way to understand this is by drawing the electric field lines. It even works if you draw it in 2D. Start with a single point. The lines come out in every direction. The electric field is strongest where the lines are closest together. Now move on to a single plate of charge. Each point on the plate emanates some electric field lines, but they get crowded together and forced to move perpendicular the plate. That's because the electric field lines aren't allowed to cross each other (because they are depictions of a vector field; only one vector is allowed at every point in space, and the vector can only point in one direction). The electric force of each charge in the plate is equal in each direction, but the forces parallel to the plate surface cancel with the opposing force of the neighboring charges. They try to push out in all directions, but only the perpendicular part comes out, since all the rest gets cancelled by the neighbors, except for charges near the edges of the plate.

    For two plates, it's basically the same as one plate, but it is more uniform and more practical in a lab.
     
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