691_Physics ProblemsTechnical Physics

# 691_Physics ProblemsTechnical Physics - Chapter 24 P24.4(a...

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Chapter 24 31 P24.4 (a) ′ = A 10 0 30 0 .. cm cm af == =′ ° =− A EA EA 300 0 030 0 7 80 10 0 030 0 180 234 4 cm m kN m C 22 2 . cos c o s . , , , Φ Φ Φ θ ej bg (b) Φ EA A , cos . cos . × ° 780 10 600 4 10.0 cm 30.0 cm 60.0Þ FIG. P24.4 Aw ° F H G I K J ° = + 30 0 30 0 10 0 60 0 600 0 060 0 7 80 10 0 060 0 60 0 2 34 4 . cos . . c o s . . , cm cm cm cm m kN m C 2 a fa f a f Φ (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 kN m C kN m C = ⋅+ + + = 2 3 4 2 3 4 000 0 . P24.5 (a) Φ E abA a A =⋅ = + ⋅ = i j i ±± ± (b) Φ E abA b A =+ ⋅= ij j (c) Φ E abA =+⋅= ijk 0 P24.6 Only the charge inside radius R contributes to the total flux. Φ E q = 0 P24.7 Φ E EA = cos through the base Φ E = 52 0 36 0 180 1 87 c o s . kN m C 2 . Note the same number of electric field lines go through the base as go through the
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## This note was uploaded on 12/14/2011 for the course PHY 203 taught by Professor Staff during the Fall '11 term at Indiana State University .

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