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Ch 24: Gauss’s Law 24.1 Electric Flux Think of flowing water as an analogy where the flow lines would be mass flow (mass/time) per area. Here E is field lines per area. We define a quantity called electric flux, E which is EA in this simple case. Electric flux is total number of lines through the area. In the water analogy we would have mass flow. 24.1 Electric Flux The total electric flux E which goes through the vertical plane A also goes through the diagonal plane. Noting that EA = EAcos, we see that E = E•A, where A is a vector normal to the area of value equal to the area, is a consistent definition for E. For small area patches dE = E•dA E = E•A = EAcosi E is positive. Outflow is positive. E is negative. Inflow is negative. E is zero. Inflow equals outflow. Ch 24: Gauss’s Law 24.2 Gauss’s Law CT1: S1 equals CT2: S2 equals CT3: S3 equals -3Q CT4: S4 equals A. -3Q/0 B. -2Q /0 C. -Q /0 D. 0 E. Q /0 F. 2Q /0 G. 3Q /0 CT5: A icosahedron has 20 equal triangular faces as pictured above. Assume a charge q is placed at the center of the icosahedron (an equal distance from each face). Using the symmetry of the situation, determine how much electric flux goes through each face. A. q/0 B. q/40 C. q/60 D. q/200 E. none of the above Gauss's law: using superposition qin is the sum of the charges enclosed by the Gaussian surface. E = E dA = qin / 0 A. Symmetries: 1) spherical 2) cylindrical (linear) 3) planar B. Method: 1) note symmetry 2) draw appropriate Gaussian surface 3) calculate electric flux E 4) set E = qin / 0 5) solve for E 24.4 Perfect Conductors in Electrostatic Equilibrium 1. E = 0 inside perfect conductors 2. The charge must reside on the surface of a perfect conductor 3. E = / 0 n where n is a unit vector normal to the surface and pointing outward. 4. is greatest at points of least radii of curvature (i.e. pointy) Before Class Assignment 2/6/08 13 1 1 correct I don’t know no explanation P24.44 (p. 689) 24.3 Application of Gauss’s Law to Various Charge Distributions CT6: At a distance r >c from the common center, the electric field is 2Q -3Q A. 2Q/40r2 B. -3Q/40r2 C. -Q/40r2 D. -Q/40r E. constant P24.39 (p. 689) CT7: The electric field in the conducting cylinder is A. 0 B. /20 C. /40 D. /20r E. /40r F. r/40 G. r/40 CT8: The electric field between the wire and the cylinder is A. 0 B. /20 C. /40 D. /20r E. /40r F. r/40 G. r/40 P24.35 (p. 688) Left 2 2 1 Gaussian Surface 1: A squat cylinder 3 3 Right 1 Gaussian Surface 2: A squat cylinder E conducting plates so on each side A. B. C. D. E. Table 24-1, p.754 Summary Gauss’s Law: E = E dA = qin/0 • Spherical Symmetry: use concentric, spherical Gaussian surface • Cylindrical Symmetry: use concentric, cylindrical Gaussian surface, assume far from ends of long thin cylinder • Planar Symmetry: use a disk shaped Gaussian surface, assume far from edges of planar surface E will be constant over the surface and either parallel or normal to the area so E dA will either be EA or 0. You may have to integrate to get qin. Homework % 9/11/2007 Average 53% Low 0% High 100% Hand-in 4pts and BCA 2pts 6 Number in 5 4 3 2 1 0 0s 10s 20s 30s 40s 50s Grade Bin 60s 70s 80s 90s