Homework #3, E&M Phy2321 9th ed. Ch. 24 P.3,4,8,10,11,13,24,27,33,35,37,39,45, [Old 8th ed. Ch. 24 P. 1,4,6,10,11,13,22,23,31,33,35,37,43 ] 3) E=4.14MN/C 4) a) phi = -2.34 kNm^2/C b) phi= +2.34 kNm^2/C c) phi= 0 8) Phi=-226 Nm^2/C 10) a) Q=-55.7 nC b) The negative charge has a spherically symmetric charge distribution, concentric with the spherical shell. 11) Phi_1 = -Q/e Phi_3 = -2Q/e Phi_2 = 0 Phi_4 = 0 13) rho = 1.77E-12 C/m^3 The charge is positive. 22) a) 16.2 MN/C b) 8.09 MN/C c) 1.62 MN/C 27) E=508 kN/C upward 33) E = (rho*r)/(2*e_0) radially away from the cylinder axis 35) a) E=0 b) E=365 kN/C c) E=1.46 MN/C d) E=649 kN/C Each is directed radially outward. 37) a) E=0 b) E=5400 N/C outward c) E=540 N/C outward 39) E=sigma/e 45) a) lambda_inner = -lambda b) lambda_outer = 3*lambda c) Gauss's law: E=6k (lambda/r) radially outward Check 22 and 35. Total score = 10 pts. [OLD CONCEPT QUESTIONS -- not assigned -- Q1. The flux of sunlight on a given area will be less when the Sun is low in the sky. The decreased flux results in colder weather. Q2. The surface must enclose a positive total charge. Q3. The net flux through any gaussian surface would be zero. This is because either: the field is uniform so the field lines entering one side will come out on the other side, or the surface would not contain any charge so, by Gauss's law, the net flux must be zero. Q4. i) Equal amounts of flux pass through eachof the size faces of the cube. (e) q/6e ii) (c) q/2e iii) (a) 0 ]