Electric Field in Matter - Electricity and Magnetism - Solved Exam, Exams of Electromagnetism and Electromagnetic Fields Theory

Download Electric Field in Matter - Electricity and Magnetism - Solved Exam and more Exams Electromagnetism and Electromagnetic Fields Theory in PDF only on Docsity! PC3231 Electricity and Magnetism 2 (Semester 1: AY 2011-12) I. Elecric eld in matter (i) Applying the Gauss's law in dielectic, D = σ in each slab. Note that D = 0 inside the metal parallel plate. (ii) Dielectric constant is another name for relative permittivity r. Us- ing D = E and  = r0, E1 = σ 20 for slab 1 and E2 = 2σ 30 for slab 2. (iii) Using P = 0χeE and χe = 1 + r, P1 = σ 2 for slab 1 and P2 = σ 3 for slab 1. (iv) Potential dierence V = E1a+ E2a = 7σa 60 . (v) Volume bound charge density ρb = 0 as the polarisation is uniform. At the top of slab 1, σb = −σ2 ; at the bottom of slab 1, σb = σ 2 . At the top of slab 2, σb = −σ3 ; at the bottom of slab 2, σb = σ 3 . II. Stress and momentum (i) Ex = 0, Ey = 0 and Ez = − σ0 ; Bx = 0, By = 0 and Bz = 0. Ele- ments of the stess tensor: Tij = 0 ( EiEj − 12δijE 2 ) + 1µ0 ( BiBj − 12δijB 2 ) , where i and j can be either x, y or z. Hence, T = σ2 20  −1 0 00 −1 0 0 0 1  (ii) As B = 0,S = 0. Thus, integrating over the xy plane: da = −dxdyẑ (negative because outward with respect to a surface enclosing the top plate). Thus, Fz = ´ Tzzdaz = − σ 2 20 A. The force per unit area is f = FA = − σ2 20 ẑ. (iii) σ 2 20 is the momentum per unit area, per unit time, crossing the xy plane. III. Rectangular waveguide (i) Let a as the length of the longer side of the rectangular waveguide. For TE10, Ez = 0, Bz = B0 cos ( πx a ) , Ex = 0, Bx = −ik (ωc ) 2−k2 π aB0 sin ( πx a ) , Ey = iω (ωc ) 2−k2 π aB0 sin ( πx a ) and By = 0. 1