Electric field due to point charge on an infinite metal plat

  • #1
Titan97
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If there is a small object with positive charge placed above a metal plate, the object induces a negative charge on the surface of the plate facing the object. Let's call this surface as S1.
(The metal plate is initially uncharged).

But from conservation of charge, the net charge in a metal plate has to be zero. So the surface below S1 will get a positive charge.
IMG_20151125_151548_447.JPG

But from the above picture, the field due to S1 and S2 are along the same direction and they don't cancel out. So how is electric field inside a conductor zero? Is it because of the electric field due to the object?
 
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  • #2
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If there is a small object with positive charge placed above a metal plate, the object induces a negative charge on the surface of the plate facing the object. Let's call this surface as S1.
(The metal plate is initially uncharged).
Ok, but not only on that surface.
But from conservation of charge, the net charge in a metal plate has to be zero. So the surface below S1 will get a positive charge.
No. The positive charge induced in the metal plate does not necessarily go there, it also distributes at the plate's borders, that is far from the plate's orthogonal axis passing through it centre (because it's farther away from the single positive charge, there) . Furthermore, to compute the field inside the material you also have to consider the one generated by the positive charge alone (you have to consider the field produced by *all* the charges).
 
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  • #3
Titan97
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So the field inside the material is because of the negative charge on S1 and that by q?
 
  • #4
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So the field inside the material is because of the negative charge on S1 and that by q?
If the plate is thin, yes. If it's not thin, that is, if the distance between the charge q and the surface S2 is of the order of the distance between q and the plate's border, then I would say that even a positive charge is induced on the surface S1.
If the metal plate is thin and has infinite extension, everything is more simple because you can compute the charge distribution with the image charge method:
https://en.wikipedia.org/wiki/Method_of_image_charges
With a plate with finite dimensions I don't know how to find the charge distribution (it doesn't mean it's not possible...).

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  • #5
Titan97
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If the plate is infinite, the field inside the conductor becomes σ/ε because if negative charge on S1 and kq/r^2 due to the object if r is the distance from q to a point inside the conductor. Will they cancel out?
 
  • #6
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The whole point of the charge separation in the metal is to cancel the field produced by the point charge. I don't mean this as an intention or goal. But the charges in the conductor will move around until this equilibrium is reached and the net field is zero everywhere you have a free charge.
So the charge distribution is such that the field is equal and opposite to the field that would be produced by the charge in that space (filled with metal) in the absence of the metal. It is not a uniform distribution of charge. The distribution of electrons will not be uniform so neither will be that of the unbalanced positive charge.
 
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  • #7
Titan97
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So what about the field outside the conductor? Will it still be σ/ε?
 
  • #8
Chandra Prayaga
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Always, irrespective of thick or thin, finite or infinite, plates or any shape, the electrostatic field inside a conductor is zero. If you have any charge outside, the surface charge on the conductor will readjust itself, and after readjustment, the field INSIDE the conductor is zero.
 
  • #9
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So what about the field outside the conductor? Will it still be σ/ε?
Certainly, immediately outside one of its surfaces it has that value and it is orthogonal to the surface, but it's not uniform, it is function of the point: σ = σ(x,y,z).

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  • #10
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Always, irrespective of thick or thin, finite or infinite, plates or any shape, the electrostatic field inside a conductor is zero. If you have any charge outside, the surface charge on the conductor will readjust itself, and after readjustment, the field INSIDE the conductor is zero.
Certainly. I specified "thin", "infinite", etc to precise what the charge distribution would be in that simple cases. But the fact E = 0 inside the conductor is always true in electrostatics.

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  • #11
Chandra Prayaga
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Certainly, immediately outside one of its surfaces it has that value and it is orthogonal to the surface, but it's not uniform, it is function of the point: σ = σ(x,y,z).

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Absolutely. And, for a point charge outside an infinite plane conductor, the distribution of the charge density is a standard problem.
 
  • #12
Titan97
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In many books, its given that if a charged infinite plate is kept in parralel with another infinite uncharged plate like this:
IMG_20151126_200502_972.JPG


the charge distribution will be like this:

IMG_20151126_200531_971.JPG


But you said there will be charges at the border as well.
Also, I have a doubt about the charge induced. Will total induced charge on one uncharged surface by another charged surface always have equal magnitude and opposite sign?
 
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  • #13
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In many books, its given that if a charged infinite plate is kept in parralel with another infinite uncharged plate like this:
View attachment 92452

the charge distribution will be like this:

View attachment 92451

But you said there will be charges at the border as well.
But do you realize that this is a totally different configuration with respect the one you posted at the beginning? There you had *a single charged object of small dimensions* over a metal plate; here instead you have an infinite metal plate in front of another...
Also, I have a doubt about the charge induced. Will total induced charge on one uncharged surface by another charged surface always have equal magnitude and opposite sign?
The charge distribution you have wrote in your drawing seems correct to me: in the region between the two metal plates the are no charges so the total flux of the electric field exiting a volume of space put there must be zero, that is if a certain amount of flux enter such a volume through one of its surfaces, the same amount of flux must exit by another. If you take a volume with a shape of a rectangular box, with two faces touching the two metal plates, you understand that the charge density in the plate at the left must be equal and opposite to the one in the plate at right (so that the flux of E entering the box at right equals the flux of E exiting at left).

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