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Two Oppositely Charged Parallel Planes, Infinite In Extent

  1. Feb 23, 2013 #1
    1. The problem statement, all variables and given/known data
    Two parallel plates having charges of equal magnitude but opposite sign are separated by 14.0 cm. Each plate has a surface charge density of 37.0 nC/m2. A proton is released from rest at the positive plate.

    (a) Determine the magnitude of the electric field between the plates from the charge density.

    (b) Determine the potential difference between the plates.

    (c) Determine the kinetic energy of the proton when it reaches the negative plate.

    (d) Determine the speed of the proton just before it strikes the negative plate.

    (e) Determine the acceleration of the proton.

    (f) Determine the force on the proton.

    (g) From the force, find the magnitude of the electric field.


    2. Relevant equations



    3. The attempt at a solution

    I have a few questions before I give you my solution to part a):

    Are we allowed to assume that the two plates are in electrostatic equilibrium? If so, why?

    If I attached a line to both planes, the line being perpendicular to both planes, the electric field will be constant across this entire line. Why is this so?

    Here is my attempt at part a), so far:

    To find the electric field at a point between the two planes, I'll apply Gauss's law. I'll use a cylinder for my Gaussian surface, where the axis of the cylinder will be parallel to the electric field. The surface area of a cylinder is, [itex]2 \pi r^2 + 2 \pi r h[/itex]; since none of the electric fields generated by the plane flows through the curved surface, there won't be any electric flux, and, consequently, we can remove the second term from our surface area equation. Additionally, since we only care about the electric field through one circular face of the cylinder, the one in between both planes, we can remove the 2 from first factor.

    So, [itex]E(\pi r^2)=\frac{q_{enc}}{\epsilon_0}[/itex]. We don't know the amount of charge enclosed by the Gaussian surface; however, we do know the area that's enclosed: it's equal to the area of the circular face of the cylinder. We also know the amount of charge contained per unit area of the plane, [itex]\sigma[/itex]. Hence, [itex]q_{enc}=\sigma A_enc[/itex]

    After substitution and simplification: [itex]E= \frac{\sigma}{\epsilon_0}[/itex]

    The exact same process can be done for the other plane. So, for the negative plane, [itex]E=\frac{- \sigma}{\epsilon_0}[/itex]. But when I add them, the result is zero. Why is that?
     
    Last edited: Feb 23, 2013
  2. jcsd
  3. Feb 23, 2013 #2

    BruceW

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    The first question - yes, you are meant to assume that the electric charges on the plates are static. This is implied by the question. You may come across questions where there is meant to be a surface current too. But since no surface currents are mentioned in the question, you can assume there aren't meant to be any.

    Also, you might be worrying that the presence of the proton will affect the charges on the plates. In reality, this will happen. But in this situation, we are meant to assume that this effect is negligible. Usually when I see this question, they specifically say that the surfaces charges are fixed in the plate. They don't specifically state that this time, but I'm pretty sure you can assume this.

    The second question - Its really just because of Gauss' law, which you have used in your answer. (Edit: you can also think of it in terms of the 'closeness' of electric field lines, if you prefer that imagery)

    You are implying that the electric field outside the two plates is zero. You are right, but why is this true?
     
  4. Feb 23, 2013 #3
    Why exactly are we permitted to assume this?

    So, because of the two infinite planes being so close, we are allowed to assume it is approximately constant, even though in reality it isn't?

    No, I was just saying that we don't have to consider it. Why is it zero?
     
  5. Feb 23, 2013 #4

    BruceW

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    Well really, they should specifically say in the question that the surfaces charges are fixed, and there are no surface currents. But from the way the question is written, it kind of looks like that is supposed to be true for this problem. (In other words, this is just about the issue of interpreting what the question is supposed to mean. It is not due to something physical about the problem that means there could not be surface currents).

    No, its got nothing to do with how close the plates are. Even if there was just one plane, think about what the field lines would look like. Can they spread out, given that the plate is infinitely wide?

    Hmm. Actually it depends on how you are doing this problem. Either you can work out the electric field due to each plate separately (using Gauss' law), and then add up their electric fields to get the actual electric field. OR you can use Gauss' law to calculate the electric field due to both the plates directly. Which way do you want to do it?
     
  6. Feb 23, 2013 #5
    Oh, I see what I was doing wrong! Yes, the electric field lines emanate perpendicularly from the infinite plane, they don't spread out at all. This is how I was thinking: suppose we have a point charge, and only one electric field line was passing through it; as you go down that single electric field line, the force decreases. But that isn't the case, is it? Does the intensity of the force depend upon how much of the electric field is acting on it?

    It seems like i'd understand the process better if I calculated the electric field due to both plates, and then add them together.
     
  7. Feb 23, 2013 #6

    BruceW

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    As I'm sure you know, the electrostatic force is the electric field times by charge of the test particle. So the magnitude of force is charge times magnitude of electric field. And if you draw some field lines, then the 'closeness' of the field lines tells you how the magnitude of the electric field varies through space. (Where the field lines are closer to each other, the field is stronger). As you said, the electric field from a point charge decreases as you move away from it. This can be explained by the way the electric field lines are getting further apart as you move away from the point charge. But of course, this won't happen for an infinite plate of charge. (which explains why the strength of the electric field is the same even when you move further away from the plate).

    Okay, I agree, doing them separately is probably most intuitive. Okay, so lets start with just the positively charged plate. As you said in your first post, the charge enclosed is [itex]\sigma \pi r^2[/itex] Also, you are right that there is no flux coming out of the curved sides. So the only way flux can come out is through the flat parts of the cylinder. (both above and below the plate). If you make your Gaussian cylinder have flat parts which are both equally far from the plate, then the two flat parts are symmetric. So how is the electric field related at these two locations? And, using Gauss' law, what is the electric field at these two locations?
     
  8. Feb 23, 2013 #7
    Well, I calculated the electric field due to the positive plane, in my first post, by considering the flux through only one face, because we don't care about the face that isn't between the two planes. And wouldn't I go through a similar process for the negative plane, getting the negative of the answer I got for the positive plane?
     
    Last edited: Feb 23, 2013
  9. Feb 23, 2013 #8

    BruceW

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    no, its not right. We are calculating the electric field due to each plate separately. So that means think about a situation where there is only a positive plate, no negative plate. So in this situation, what is the electric field either side of the plate?
     
  10. Feb 23, 2013 #9
    Would it be, [itex]E= \frac{\sigma}{2 \epsilon_0}[/itex]?
     
  11. Feb 23, 2013 #10

    BruceW

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    exactly :) And it will be pointing away from the plate. Now you can use similar calculation for the negative plate, and finally add up their electric fields in the 3 regions: inside the plates, above the plate, below the plates.
     
  12. Feb 24, 2013 #11
    Yes, but wouldn't it be the exact same calculation, the difference being that it is negative? And if I added the two electric fields, wouldn't they sum to zero?
     
  13. Feb 24, 2013 #12

    BruceW

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    they do sum to zero in some regions, but not others. Draw it out, with electric field lines coming from each of the plates. (remember the direction they should point in).
     
  14. Feb 25, 2013 #13
    What do you mean they sum to zero in some regions? Inside, between the two plates, I know that the electric field is not suppose to be zero, but when I do sum the two electric field contributions, its zero. And how could it be zero outside the region between the two plates, there is no way for the fields to interact--the infinite planes divide space.
     
  15. Feb 25, 2013 #14
    Could someone possibly help? BruceW and I appear to be going in circles.
     
  16. Feb 25, 2013 #15

    Doc Al

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    Let's say that the negative charged plate (at x = 0) is to the left of the positive plate (at x = 14.0 cm) and that both plates are in the y-z plane. (So the fields will be along the x-axis.)

    Start with the negative plate. Which way does its field point:
    (a) in the region to the left of the plate (x < 0)
    (b) in the region to the right of the plate (x > 0)

    Answer the same questions for the positive plate. Which way does its field point:
    (a) in the region to the left of the plate (x < 14.0 cm)
    (b) in the region to the right of the plate (x > 14.0 cm)

    Then for the region between the plates, see how the two field contributions add up.

    As BruceW advises, draw yourself a picture.
     
    Last edited: Feb 25, 2013
  17. Feb 25, 2013 #16
    For the negative plate:

    (a): The electric field lines point towards the negative plate
    (b): The electric field lines originate on the inner surface of the negative plate and point towards the inner surface of the negative plate

    For the positive plate:

    (a): Same answer given in part (b) for the negative plate
    (b): The electric field lines emanate away from the outer surface of the positive plate

    So is the electric field due to the positive plate negative, because it points in the negative direction? What about the negative plate? It's already negative and it points in the negative direction, so that would be a double negative and still result in zero when you sum the two electric fields.
     
  18. Feb 25, 2013 #17

    Doc Al

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    OK. For a negative plate, the field points toward the plate. So:
    (a) to the right
    (b) to the left

    For a positive plate, the field points away from the plate. So:
    (a) to the left
    (b) to the right


    Yes, between the plates the field from each points to the left (thus is negative).
    Double negative? You are adding, not multiplying. What happens when you add two negative numbers?
     
  19. Feb 25, 2013 #18
    Isn't the electric field generated by a negative charge distribution negative? If so, it is also pointing in the negative direction. So electric field generated by the negative plate is negative and is point in the negative direction; that is how I came up with the double negative. So, the electric field of the negative plate is actually positive, and the electric field is actually negative because it is pointing in the negative direction. Add them together, the result is zero. Do you see what I am having trouble with?
     
  20. Feb 25, 2013 #19

    Doc Al

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    "Negative" just indicates, via sign convention, the direction of the field. So on one side of the plate the field would be negative, but on the other side it will be positive. (The field from a negative charge points inward toward the charge.)

    Right. In the region between the plates, the field from the negative plate is negative (meaning: It points to the left).
    We've established that the field from the negative plate is negative in the region between the plates. So what's the field from the positive plate in that region?

    Draw yourself a diagram! Draw in the direction that the field point in in all regions. Vectors only cancel when they point in opposite directions.
     
  21. Feb 25, 2013 #20
    Oh, so the electric fields sign doesn't depend on the source charge, whether it's positive or negative, but the direction?
     
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