Simple Nature - Light and Matter

corners to the disk and transform it into a square. Each corner’scontribution to the field measured at the center is canceled by thefield due to the corner diagonally across from it.What if the flea goes on a trip away from the center of the disk?The perfect cancellation of the “horizontal” fields contributed bydistant charges will no longer occur, but the “vertical” field (i.e., thefield perpendicular to the surface) will still be E ⊥ = 2πkσ, where σ isthe local charge density, since the distant charges can’t contribute tothe vertical field. The same result applies if the shape of the surfaceis asymmetric, and doesn’t even have any well-defined geometriccenter: the component perpendicular to the surface is E ⊥ = 2πkσ,but we may have E ‖ ≠ 0. All of the above arguments can be mademore rigorous by discussing mathematical limits rather than usingwords like “very small.” There is not much point in giving a rigorousproof here, however, since we will be able to demonstrate this factas a corollary of Gauss’ Law in section 10.6. The result is as follows:At a point lying a distance z from a charged surface, the componentof the electric field perpendicular to the surface obeyslim E ⊥ = 2πkσ ,z→0where σ is the charge per unit area. This is true regardless of theshape or size of the surface.j / Example 13.The field near a point, line, or surface charge example 13⊲ Compare the variation of the electric field with distance, d, forsmall values of d in the case of a point charge, an infinite line ofcharge, and an infinite charged surface.⊲ For a point charge, we have already found E ∝ d −2 for themagnitude of the field, where we are now using d for the quantitywe would ordinarily notate as r. This is true for all values of d,not just for small d — it has to be that way, because the pointcharge has no size, so if E behaved differently for small and larged, there would be no way to decide what d had to be small orlarge relative to.For a line of charge, the result of example 10 isE =kλLd 2√ 1 + L 2 /4d 2 .In the limit of d ≪ L, the quantity inside the square root is dominatedby the second term, and we have E ∝ d −1 .Finally, in the case of a charged surface, the result is simply E =2πσk, or E ∝ d 0 .Notice the lovely simplicity of the pattern, as shown in figure j. Apoint is zero-dimensional: it has no length, width, or breadth. Aline is one-dimensional, and a surface is two-dimensional. As the580 Chapter 10 Fields