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A point is stronger than a plane?

  1. Nov 26, 2009 #1
    If you shoot a charge +q at a plane with charge +Q, you will eventually be able to reach it, since E is constant.
    But if you shoot a charge +q at another charge +q, the E field increases as you approach the point charge, so you will never be able to reach it.

    How can it be that a point charge is more powerful than a plane, even though a plane is made up of point charges?

    Thakns
     
  2. jcsd
  3. Nov 26, 2009 #2
    That the plane creates a constant electric field is just an aproximation. As you get closer, you will start feeling the inhomogeneities, and the charge will either go through holes between particles (scattered, of course), or will bounce back in case it happened to hit another charge
     
  4. Nov 26, 2009 #3
    Can you please explain what inhomogeneities you mean? Do you mean the charge is quantized instead of perfectly uniformly distributed?
     
  5. Nov 26, 2009 #4
    Can you specify (1) how +Q is distributed on the plane, (eg. is it a uniform per unit area density?) and (2) whether the plane is conducting?
     
  6. Nov 26, 2009 #5

    rcgldr

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    Assuming the plane has some fixed charge per unit area, as the charged object gets very close to the plane, most of charges beyond some small radius nearly cancel each other out, only contributing some small amount to the total electrical force perpendicular to the plane. Effectively the object is approaching some tiny portion of the plane, and that tiny portion has only a tiny charge. As the object gets closer to the surface of the plane, the effective opposing charge decreases, so that the opposing force remains constant.

    With a point source, all of the charge resides in a single point, so no matter how close the charged object approaches, the opposing amount of charge remains constant, and the opposing force increases by a factor of 1 / r2.
     
    Last edited: Nov 27, 2009
  7. Nov 27, 2009 #6
    I go only partly along with this explanation. Don't forget that a +ve point charge will induce a -ve point charge in the plate, called a mirror image. Charge and mirror image will attract each other just like real charges do. At short distance this force will become dominant, and the +ve charge will slam into the plate.
     
  8. Nov 27, 2009 #7

    Redbelly98

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    This is true if the plane is a conductor, but that was never a given here. Note Jeff Reid is talking about the case where "the plane has some fixed charge per unit area", which implies a non-conducting surface. An image charge would make it a varying charge-per-area on the surface.
     
  9. Nov 28, 2009 #8
    You make a valid point. However I've got a sneaky feeling that the op was talking about a conducting plane.
     
  10. Dec 5, 2009 #9
    Redbelley posts:
    I don't like that explanation, nor any of the others for that matter: as a POINT charge gets arbitarily close to any charged plane,or another charge, doesn't the force of repulsion become arbitarily large??

    GATO suggests the constant electric filed is an approximation; so what is an exact expression??
     
  11. Dec 5, 2009 #10

    rcgldr

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    As posted before, since the charge per unit area, σ, is assumed constant, then the force remains constant, regardless of position, except when the point charge is on the plane (zero force) or goes through it (direction of force changes). The forces parallel to the plane cancel, and the force perpendicular to the plane is constant because as the particle approaches, the angle between the point charge and plane from the charged areas on the plane decreases. As the charge approaches the surface of the plane the effective opposing charge decreases, resulting in a constant force. Here is a link to a derivation based on infinitely thin rings used to form a disk. The other method is to use infinetly long and thin rectangles to form a plane. For the disc case, as R -> , then

    E = 2 π k σ

    http://hyperphysics.phy-astr.gsu.edu/%E2%80%8Chbase/electric/elelin.html#c3 [Broken]

    Another example would be a finite sized plane or disk with finite charge. At large distances, the disk ends up with about the same field as a point charge, but at small distances, a point charge can only get very close to a small fraction of the total surface of the plane. For example if a point charge approaches the center of a disk with a radius of 1 meter, the point charge can never get closer than 1 meter from the edge of the disk. As the point charge gets closer, only a small area and corresponding small component of charge of the plane or disk near the charge has a significant contribution to the force on the point charge, and as the point charge approaches the surface of the plane or disk, the opposing force approaches a constant value, not ∞.

    Since I can't find the rectangle based derivation, here is a summary.

    For an infinite line charge, E = 2 k λ / r, where λ is charge per unit length and r is distance from the line. Here is the derivation:
    http://hyperphysics.phy-astr.gsu.edu/%E2%80%8Chbase/electric/elelin.html#c1 [Broken]

    For an infinite plane case, infinitely long strips (rectangles) approximate a line as their width -> 0. The area of a strip of lengh L is L dx, and the charge dq = σ L dx. The charge per unit length dλ = dq/L = σ dx. Assume the plane exists on the x-y plane, then the magnitude of the field at any point in space from a strip is dE = 2 k σ / r, where r is the distance from a strip to that point in space. In the case of the entire plane, the x components cancel because of symmetry, with only a net force in the z direction, and for each strip of the plane, dEz = dE (z / r) = dE sin(θ).

    [tex]E = 2 k \sigma \int_{-\infty}^{+\infty} \frac{{sin}(\theta)dx}{r}[/tex]

    [tex]{sin}(\theta) = z / r[/tex]

    [tex]E = 2 k \sigma \int_{-\infty}^{+\infty} \frac{z {dx}}{r^2}[/tex]

    [tex]r^2 = z^2 + x^2[/tex]

    [tex]E = 2 k \sigma z \int_{-\infty}^{+\infty} \frac{dx}{z^2 + x^2}[/tex]

    [tex]E = 2 k \sigma z \left [ \frac{1}{z} tan^{-1}\left (\frac{x}{z}\right )\right ]_{-\infty}^{+\infty} [/tex]

    [tex]E = 2 k \sigma z \left ( \frac{1}{z} \right) \left ( \frac{+ \pi}{2} - \frac{- \pi}{2} \right ) [/tex]

    [tex]E = 2 \pi k \sigma [/tex]
     
    Last edited by a moderator: May 4, 2017
  12. Dec 5, 2009 #11
    I think the answer is in your question. If on the one hand you assume a plane is made up of point charges, then it must also have spaces in between. Qualitatively, the spaces make the plane weaker allowing particles to find a way through the field. Now, that only matters when you get close enough to see the spaces. Far away where the point charge picture fades, qualitatively, why is it weaker than a point charge?

    I ask, wouldn't you rather get hit by a pillow than a ball bearing? If they weigh the same why would you prefer the pillow, especially if the pillow is actually bigger?
     
  13. Dec 5, 2009 #12
    Rutherford's gold foil experiment exposes some of the logic applied on this thread.

    This is a short shockwave flash explaination of shooting alpha particles at gold foil:

    http://www.mhhe.com/physsci/chemistry/essentialchemistry/flash/ruther14.swf

    Two postive charges cannot occupy the same space. We think of a point charge with zero volume, like a point mass, but this is an idealization. On the atomic scale charge is discrete and widely spaced.

    There is still a good question here about the strength of the E field and how it influences the kinetic energy of the incident particles. The question of whether a metal or dielectric surface, etc.
     
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