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695_Physics ProblemsTechnical Physics

# 695_Physics ProblemsTechnical Physics - Chapter 24(a E=(b...

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Chapter 24 35 P24.24 (a) E k Qr a e = = 3 0 (b) E k Qr a e = = × × = 3 9 6 3 8 99 10 26 0 10 0 100 0 400 365 . . . . e je j a f a f kN C (c) E k Q r e = = × × = 2 9 6 2 8 99 10 26 0 10 0 400 1 46 . . . . e je j a f MN C (d) E k Q r e = = × × = 2 9 6 2 8 99 10 26 0 10 0 600 649 . . . e je j a f kN C The direction for each electric field is radially outward . *P24.25 mg qE q q Q A = = F H G I K J = F H G I K J σ 2 2 0 0 Q A mg q = = × × = 2 2 8 85 10 0 01 9 8 0 7 10 2 48 0 12 6 . . . . . e j a fa f C m 2 µ P24.26 (a) E k r e = 2 λ 3 60 10 2 8 99 10 2 40 0 190 4 9 . . . . × = × e j b g Q Q = + × = + 9 13 10 913 7 . C nC (b) E = 0 *P24.27 The volume of the spherical shell is 4 3 0 25 0 20 3 19 10 3 3 2 π . . . m m m 3 a f a f = × . Its charge is ρ V = − × × = − × 1 33 10 3 19 10 4 25 10 6 2 8 . . . C m m C 3 3 e je j . The net charge inside a sphere containing the proton’s path as its equator is ×
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