Ch 24 Gausss Law 24 1 Electric Flux

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 d E = E • d. A

E = E • A = E Acos i

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

CT 1: S 1 equals CT 2: S 2 equals CT 3: S 3 equals -3 Q CT 4: S 4 equals A. -3 Q/ 0 B. -2 Q / 0 C. -Q / 0 D. 0 E. Q / 0 F. 2 Q / 0 G. 3 Q / 0

CT 5: 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/4 0 C. q/6 0 D. q/20 0 E. none of the above

Gauss's law: using superposition qin is the sum of the charges enclosed by the Gaussian surface. E = = 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

P 24. 44 (p. 689)

24. 3 Application of Gauss’s Law to Various Charge Distributions CT 6: At a distance r >c from the common center, the electric field is 2 Q -3 Q A. 2 Q/4 0 r 2 B. -3 Q/4 0 r 2 C. -Q/4 0 r 2 D. -Q/4 0 r E. constant

P 24. 39 (p. 689)

CT 7: The electric field in the conducting cylinder is A. 0 B. /2 0 C. /4 0 D. /2 0 r E. /4 0 r F. r/4 0 G. r/4 0

CT 8: The electric field between the wire and the cylinder is A. 0 B. /2 0 C. /4 0 D. /2 0 r E. /4 0 r F. r/4 0 G. r/4 0

P 24. 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 = = 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 will either be EA or 0. You may have to integrate to get qin.