Gauss Law For Dielectrics Give D 9 4tr2 In All Three Regions And Symmetry Requires Do Do 0 Using E D D 1 (107.86 KiB) Viewed 32 times
Gauss' law for dielectrics give D, = 9/4tr2 in all three regions and symmetry requires Do = Do = 0. Using E, = D,/€ = D./€(1 + Xe) gives 9 1 r>c 4TE 2 E = 9 1 4πε γι2 b<r<c 91 r<b 4πε, r2 Check: As Xe → 00, the electric field in the dielectric is zero and the other electric fields are as expected from Gauss' law. As Xe →0. the field in all regions is that for a point charge. Using (r) - 0(0) = - 5 E.(dir" and "(Qc) = 0 can be used with the E. for r > c to find ø(r) for r>c. Ex(r"}dr" can be used to find € [") for 6 <r<c, where 4(e) is known from the potential found for r>c. Then (r) - V(C) = The potnetial for r <b can be found in a similar way. The result is 9 1 4πε, Τ2 r>C 9 1 + b<r<c 4πε η l+Xe (3-3)] ( 2)] q 1 41€. T + 1 1 + x r<b Activate Windows Go to Settings to activate Windows The previous two checks apply. In addition, we expect continuity to apply, so the potentials should match at their boundaries.
10.2 Long Cylindrical Shell Containing Magnetizable Material A long thick cylindrical shell with inner radius R; and outer radius R has a magnetic susceptibility of Xm and is initially unmagnetized. The cylinder is centered on the origin and aligned with the z-axis. There is a long wire that carries current I, that runs along the s-axis in the +: direction. 1. The total field is given by B =Bext + Bj. where B, is the field due to K, and J, that are induced by the external magnetic field due to the current-carrying wire. 1. Use an analog of Approach 1. in HW 7.3.1 to show that B = (1 + Xm)Bezt inside the cylinder. That is, start with fH. dl = If end. 2. Use an analog of Approach 3. in HW 7.3.1 to show that B = (1 + Xm)Bezt inside the cylinder. That is, use M = (Xm/(1 + Xm))B/Ho, the equations for Kb. Jb, and $ B. dl = Molencl. Activate Windows Go to Settings to activate Wind
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