Distance from plane?

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Distance from plane??

When you have infinite plane and you check the electric field in a point near it, you find out that it doesn't depend on the distance of this point from the plane.
Why the distance from the plane does not affect the electric field produced by it??
and if you could give me an concrete example of why its happening.

thanks
 

Answers and Replies

  • #2
rcgldr
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All the forces parallel to the plane cancel, so the only net force is perpendicular to the plane. The plane has a uniform charge per unit area, and as you get futher away, the angle from the charges further away becomes more perpendicular, increasing their net effect, so the force is independent of distance. You can experience a similar effect in a crowded noisy room, where the sound level is about the same if your distance from the crowd is small compared to the size of the crowd. For the mathematical explanations:

The math for field strength from a solid disc:

http://hyperphysics.phy-astr.gsu.edu/Hbase/electric/elelin.html#c3

as [tex]R \ \rightarrow \infty [/tex]

then [tex] \frac{z} {\sqrt{z^2 + R^2}} \ \rightarrow \ 0 [/tex]

and you end up with [tex]E_z = k \ \sigma \ 2 \ \pi [/tex]

An alterative approach is to consider the field from an infintely long line (= 1/z), then integrate an infinitely large plane composed of infintely long rectangles that approach infinitely long lines as their width approaches zero.

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]
 
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  • #3
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This is just because the plane is infinite. For real surfaces it will depend on distance.
For infinite plane, as the point goes further away, the contributions from regions far away to the normal component increase and compensate for increased distance.
 
  • #4
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The plane has a uniform charge per unit area, and as you get futher away, the angle from the charges further away becomes more perpendicular, increasing their net effect, so the force is independent of distance. You can experience a similar effect in a crowded noisy room, where the sound level is about the same if your distance from the crowd is small compared to the size of the crowd.
I didnt get the whole thing with the angles... and I dont care about the mathematical explanations, it's just dont make sense that I go further and there is same force...
Can you explain it little more clearley?
THX
 
  • #5
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If you can "understand" the idea of an infinite plane, then you should not have a problem with the constant force.
The explanation is mathematical, by the way. Even if it's in words.
 
  • #6
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but why isnt it depends on the distance?? it doesnt make any sense... if you get further the force need to get smaller...
 
  • #7
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someone??
 
  • #8
rcgldr
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Maybe this will help. Look at your monitor and move closer to it, the image doesn't get any brighter (the edges might get darker if it's a certain type of LCD monitor, you won't get that effect from a CRT monitor, but I doubt you find one to try this with). Unless you are really close to the monitor, you can see beyond the edges of the monitor, and in a dark room, the total light you'd see would only exist with the boundaries of the monitor. If you were to go very far away from the monitor, eventually it would look like a dim point source of light. However if the monitor was huge, like a movie screen, you'd have to get much further away before that happens. If the monitor was infinite in size, there would be no edges to see and the brightess would be the same no matter how far away you are.
 

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