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Extra resources for Companion to J.D. Jackson's Classical Electrodynamics
A1 = κ−1 2 b E0 κ+1 −2E0 1+κ B−1 = 0 B1 = D1 = −4E0 (1 + κ)2 For the cylindrical cavity, I place the surface of the outer shell at infinity, b → ∞. In this limit A1 is ill-defined, so we’ll ignore it. 10 Two concentric conducting spheres of inner and outer radii a and b, respectively, carry charges ±Q. The empty space between the spheres is half-filled by a hemispherical shell of dielectric (of dielectric constant κ = ǫǫ0 ), as shown in the figure. a. Find the electric field everywhere between the spheres.
Biot-Savart’s law tells us how to find the magnetic field at some point P (r) produced by a wire element at some other point P2 (r ′ ). At P (r): dB = µ0 r − r′ Idℓ′ × 4π |r − r ′ |3 The total B-field at a point P is the sum of the dB elements from the entire loop. So we integral dB around the closed wire loop. B= dB = µ0 I 4π Γ dℓ′ × r − r′ |r − r ′ |3 There is a form of Stokes’ theorem which is useful here: dℓ′ × A = dS ′ × ∇′ × A. I’ll look up a definitive reference for this someday; this maybe on the inside cover of Jackson’s book.
Set µ1 = 1 and µ2 = µ. Notice that all the signs are positive. For the components of the current parallel to the surface, this is exactly as expected. For the z component, we have a reflection of a reflection or simply a weakened version of the original current as our image; therefore, the sign is positive. J ∗∗ = 2µ J µ+1 To get a better understanding of the physics involved here, I will derive these results using the boundary conditions. We are solving ∇×H =J Which has a formal integral solution H= µ0 4πµ d3 r ′ J(r ′ ) × |r − r ′ | |r − r ′ |3 But our Js are point currents, that is J ∝ δ(r ′ − a), so we can do the integral and write 1 H ∝ I × (r − a′ ) µ 39 ˆ The cross product is what causes all the trouble.