693_Physics ProblemsTechnical Physics

693_Physics ProblemsTechnical Physics - Chapter 24 P24.15...

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Chapter 24 33 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 kQ R e 2 radially outward (normal to the surface). Therefore, the flux is this field strength times the area of half a sphere: Φ Φ curved local hemisphere curved =⋅= = F H G I K J F H G I K J = = + z EA dE A k Q R RQ Q e 2 2 00 1 2 4 1 4 2 2 π af Q 0 FIG. P24.15 (b) The closed surface encloses zero charge so Gauss’s law gives ΦΦ curved flat += 0o r flat curved =− = Q 2 0 . *P24.16 Consider as a gaussian surface a box with horizontal area A , lying between 500 and 600 m elevation. ⋅= z d q 0 : ++ −= 120 100 100 0 NC m bg a f AA A ρ = ×⋅ 20 8 85 10 100 177 10 12 12 C N m m Cm 22 3 ej . . The charge is positive , to produce the net outward flux of electric field. P24.17 The total charge is Qq 6 . The total outward flux from the cube is
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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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