aho02_EfromPtSrcIntegral

# aho02_EfromPtSrcIntegral - Physics 1B E-Field from a...

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Physics 1B: E-Field from a Point-Source Integral E-Field of a Plane of Charge 1 Problem Using the remote-point-source expression for the electric ﬁeld (rather than Gauss’s law), solve for the electric ﬁeld everywhere surrounding an inﬁnite plane with surface charge density σ . 2 Quick Solution By symmetry, E only points toward or away from the plane, and it can only depend on z . E ( z ) = K e Z distr. d q 0 R 2 ˆ R = K e Z distr. d q 0 R 3 R = K e Z plane d q 0 R 3 R z ˆ z = K e Z 0 σ 2 πs 0 d s 0 h s 0 2 + z 2 i 3 / 2 z ˆ z = 2 πK e σz ˆ z Z 0 s 0 d s 0 h s 0 2 + z 2 i 3 / 2 = 2 πK e σz ˆ z Z z 2 1 2 d u u 3 / 2 = 2 πK e σz ˆ z ± - 1 u 1 / 2 ² z 2 = 2 πK e σz ˆ z " 0 - ³ - 1 | z | ´ # = ± 2 πK e σ ˆ z (+ sign for z > 0 , - sign for z < 0) = ± 2 π ³ 1 4 πε 0 ´ σ ˆ z = ± σ 2 ε 0 ˆ z We see that the electric ﬁeld in fact does not depend on z after all. E = σ 2 ε 0 ˆn , where ˆn points away from the plane, E = 0 inside the plane. 1

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3 Explained Solutions 3.1 Qualitative Analysis As with every problem, we should make a qualitative analysis before getting quantitative; we can get a sense of what the answer should be and potentially simplify the calculation with our conceptual ideas. A good way to start is to draw a picture: We can arbitrarily choose to draw the plane in a horizontal orientation. Let’s pick a random point P above the plane and see if we can reason what direction the electric ﬁeld (“E-ﬁeld”) E P at this point is pointing. Let’s assume for now that the charge density is positive, σ > 0. If we examine a small patch on the plane, it will have a small positive amount of charge d q 1 and that will contribute a small amount of E-ﬁeld d E 1 P at the point P that points directly away from the patch. (See the illustration on the left, below.) If we look at another patch that is symmetrically placed on the other side of P , it will contribute its own small amount of E-ﬁeld d E 2 P , and the magnitudes of these two contributions are the same, d E 1 P = d E 2 P , since the source charges d q 1 and d q 2 are the same distance from point P . The geometry of this situation is such that the horizontal components of these two E-ﬁeld vectors cancel, but the vertical components add to point away from the plane. Since an inﬁnite plane can be completely constructed from these pairs of patches, it follows that the total electric ﬁeld at point P should have no horizontal component and only a vertical component pointing away from the plane. This same logic allows us to conclude that for negative charge density, σ < 0, the electric ﬁeld must point directly toward the plane, with no component parallel to the plane. (See the illustration on the right, above.) Another kind of argument gives us this same result. This other argument is a proof by contra- diction. Here it is: Since the source charge distribution is symmetrical with respect to translation
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aho02_EfromPtSrcIntegral - Physics 1B E-Field from a...

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