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HWCh24Solutions

# HWCh24Solutions - PHYSICS 133 HOMEWORK SOLUTIONS CHAPTER 24...

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PHYSICS 133 HOMEWORK SOLUTIONS CHAPTER 24 P24.4 (a)    = 10.0 cm ( 29 30.0 cm ( 29    = 300 cm 2 = 0.030 0 m 2 Φ E , = E cos θ Φ E , = 7.80 × 10 4 ( 29 0.030 0 ( 29 cos 180 ° Φ E , = - 2.34  kN m 2 C (b)    Φ E , A = EA cos θ = 7.80 × 10 4 (29 A ( 29 cos 60.0 ° 10.0 cm  30.0 cm 60.0? FIG. P24.4    A = 30.0 cm ( 29 w ( 29 = 30.0 cm ( 29 10.0 cm cos 60.0 ° = 600 cm 2 = 0.060 0 m 2 Φ E , A = 7.80 × 10 4 ( 29 0.060 0 ( 29 cos60.0 °= + 2.34  kN m 2 C (c) The bottom and the two triangular sides all lie  parallel  to  E , so     Φ E = 0   for each of these. Thus,    Φ E , total =- 2.34  kN m 2 C + 2.34  kN m 2 C + 0 + 0 + 0 = 0 . P24.11    Φ E = q in 0 Through    S 1    Φ E = - 2 Q + Q 0 =- Q 0 Through    S 2    Φ E = + Q - Q 0 = 0 Through    S 3    Φ E = - 2 Q + Q - Q 0 =- 2 Q 0 Through    S 4    Φ E = 0 P24.15 (a) With  δ  very small, all points on the hemisphere are  nearly at a distance  R  from the charge, so the field  everywhere on the curved surface is     k e Q R 2  radially  outward (normal to the surface). Therefore, the flux is

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HWCh24Solutions - PHYSICS 133 HOMEWORK SOLUTIONS CHAPTER 24...

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