Jackson 1.3 Homework Problem Solution
KALMAN KNIZHNIK - SOLUTIONS TO JACKSON - CHAPTER 1 JACKSON 1.1 Part a Since a conductor has no E field inside, by Gauss's law ∇ · E = ρ ɛ 0 there can also be no charge density inside the conductor. This means that excess charge must lie entirely on its surface. ✷ Part b You can place a Gaussian surface just inside the conductor. Since there is no charge density inside, by Gauss's law (perhaps more easily seen in integral form) ∮ E · da = Qenc ɛ 0 there is also no E field inside, in spite of all of the charges outside the conductor. However, if there are charges inside the hollow conductor, the same law tells you that Q enc ≠ 0, meaning that the field outside is nonzero too. Hence, it shields its interior from charges outside but does not shield its exterior from charges inside. ✷ Part c The E field on the surface of the conductor must be normal to its surface, since otherwise charge would flow on the surface (we are making the assumption that the charge is static. As for the magnitude, we use Gauss's law in integral form: ∮ E · da = Q enc ɛ 0 (1) Now, dQ enc = σˆn · da, since the only charge that exists is on the surface. Thus, we rewrite Gauss's law as ∫ ∫ ∫ σ E · da = ˆn · da = (E − σ ˆn) · da = 0 (2) ɛ 0 ɛ 0 and since this must hold for any area, we conclude JACKSON 1.2 Starting with <strong>Jackson</strong>'s definition of the delta function: E = σ ɛ 0 ˆn ✷ (3) D(α; x, y, z) = (2π) 3/2 α −3 exp[− x2 + y 2 + z 2 2α 2 ] (4) we note that D → 0 unless x, y, z → 0 as well (i.e. replace x, y, z with dx, dy, and dz). Now notice that dx 2 + dy 2 + dz 2 ≡ ds 2 is the arc length squared. We can write D ∝ exp[−ds 2 /2α 2 ]. In the general coordinate system given by <strong>Jackson</strong>, we can write the square of the arc length as: ds 2 = ( du U )2 + ( dv V )2 + ( dw W )2 (5) This is not to say that dx = du/U. Rather, I have replaced the entire arc length ds, which is the same in all coordinate systems. Let du → u − u ′ , dv → v − v ′ , and dw → w − w ′ . We obtain: D(α; u, v, w) = e−(u−u′ ) 2 /2α 2 U 2 √ 2πα e −(v−v′ ) 2 /2α 2 V 2 √ 2πα e −(w−w′ ) 2 /2α 2 W 2 √ 2πα (6) This is starting to look like the usual definition of the delta function. Now for some trickery. We will replace the α's in each term with α u /U, α v /V, α w /W . We justify this by saying that each of the α's go to zero, so the replacement does not change the limit of the function. The result of this replacement is D(α; u, v, w) = e−(u−u′ ) 2 /2α 2 u √ 2παu e −(v−v′ ) 2 /2α 2 v √ 2παv e −(w−w′ ) 2 /2α 2 w √ 2παw UV W (7)
Jackson 1.3 Homework Problem Solution
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