# Ch22Word - Chapter 22 p. 1 CHAPTER 22 Gauss's Law 1....

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Chapter 22 p. 1 CHAPTER 22 – Gauss’s Law 1. Because the electric field is uniform, the flux through the circle is Φ = ? E · d A = E · A = EA cos θ . ( a ) When the circle is perpendicular to the field lines, the flux is Φ = EA cos = EA = (5.8 × 10 2 N/C)p(0.15 m) 2 = 41 N · m 2 /C . ( b ) When the circle is at 45° to the field lines, the flux is Φ = EA cos = EA = (5.8 × 10 2 N/C)p(0.15 m) 2 cos 45° = 29 N · m 2 /C . ( c ) When the circle is parallel to the field lines, the flux is Φ = EA cos = EA = (5.8 × 10 2 N/C)p(0.15 m) 2 cos 90° = 0 . 2. Because the electric field is radial, it is perpendicular to the spherical surface just beyond the Earth’s surface. The field is also constant, so the flux through the sphere is Φ = ? E · d A = – EA = – (150 N/C)4p(6.38 × 10 6 m) 2 = – 7.7 × 10 16 N · m 2 /C . Note that E and d A are in opposite directions. 3. All field lines enter and leave the cube, so the net flux is Φ net = 0 . We find the flux through a face from Φ = ? E · d A . There are two faces with the field lines perpendicular to the face, say one at x = 0 and one at x = ¬ . Thus for these two faces we have Φ x = 0 = – EA = – (6.50 × 10 3 N/C) ¬ 2 = – (6.50 × 10 3 N/C) ¬ 2 ; Φ x = ¬ = + EA = + (6.50 × 10 3 N/C) ¬ 2 = + (6.50 × 10 3 N/C) ¬ 2 . For all other faces, the field is parallel to the face, so we have Φ all others = 0 . 4. ( a ) Because the angle between the electric field and the area varies over the surface of the hemisphere, it would appear that we find the flux by integration. We see that the same flux must pass through the circular base of the hemisphere, where the field is constant and perpendicular to the surface. T h u s w e h a v e Φ = EA = E p R 2 . ( b ) If E is perpendicular to the axis, every field line must enter and leave the surface, so we have Φ = 0 . 5. The total flux is depends only on the enclosed charge: Φ = Q / Å 0 , or Q = Å 0 Φ = (8.85 × 10 –12 C 2 /N · m 2 )(1.45 × 10 3 N · m 2 /C) = 1.28 × 10 –8 C = 12.8 nC . E R d A axis

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Chapter 22 p. 2 6. The net flux through each closed surface is determined by the net charge inside. Thus we have Φ 1 = ( Q – 3 Q )/ Å 0 = – 2 Q / Å 0 . Φ 2 = ( Q – 3 Q + 2 Q )/ Å 0 = 0 . Φ 3 = (– 3 Q + 2 Q )/ Å 0 = Q / Å 0 . Φ 4 = (0)/ Å 0 = 0 . Φ 5 = (+ 2 Q )/ Å 0 = + 2Q/ Å 0 . 7. The total electric flux through the surface depends only on the enclosed charge: Φ = ı E · d A = Q / Å 0 . The only contributions to the integral are from the faces perpendicular to the electric field. Over each of these two surfaces, the magnitude of the field is constant, so we have Φ = E ¬ A E 0 A = ( E ¬ E 0 ) A = Q / Å 0 ; (410 N/C – 560 N/C)(30 m) 2 = Q /(8.85 × 10 –12 C 2 /N · m 2 ), which gives Q = – 1.2 × 10 –6 C = – 1.2 µ C . 8. The total electric flux through the surface depends only on the enclosed charge: Φ = ı E · d A = Q / Å 0 . Because the charge is at the center of the cube, we know from symmetry that each of the six faces has the same flux through it: Φ face = (1/6) Φ total = Q /6 Å 0 . 9. If we construct a spherical Gaussian surface just outside the ball, we have Φ = ı E · d A = Q / Å 0 .
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## This note was uploaded on 02/05/2010 for the course PHYSICS 37,38,39 taught by Professor Fielding during the Spring '09 term at Los Angeles Valley College.

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Ch22Word - Chapter 22 p. 1 CHAPTER 22 Gauss's Law 1....

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