ch24-p066 - 66. Since the charge distribution is...

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(c) For R 2 > R 1 > r , the enclosed charge is zero. Thus, E = 0. The electric potential may be obtained using Eq. 24-18: Vr Vr Erd r r r b g b g b g ′ = z . (d) For r = 4.00 m > R 2 > R 1 , we have () 92 2 6 6 3 12 0 (8.99 10 N m C )(2.00 10 C 1.00 10 C) 6.74 10 V. 4 (4.00 m) qq Vr r πε −− + ×⋅ × + × == = × (e) For r = 1.00 m = R 2 > R 1 , we have 2 6 6 4 0 (8.99 10 N m C )(2.00 10 C 1.00 10 C) 2.70 10 V. 4 (1.00 m) r + × + × = × (f) For R 2 > r = 0.700 m > R 2 , 66 2 02 4 1 2.00 10 C 1.00 10 C (8.99 10 N m C ) 4 0.700 m 1.00 m 3.47 10 V. rR ⎛⎞ ×× =+ = × + ⎜⎟ ⎝⎠ 66. Since the charge distribution is spherically symmetric we may write enc 0 1 , 4 q Er r = where q enc is the charge enclosed in a sphere of radius r centered at the origin. (a) For r = 4.00 m, R 2 = 1.00 m and R 1 = 0.500 m, with r > R 2 > R 1 we have 2 6 6 3 22 0 (8.99 10 N m C )(2.00 10 C 1.00 10 C) 1.69 10 V/m. 4 (4.00 m) r + × + × = × (b) For R 2 > r = 0.700 m > R 2 2 6 4 1 0 (8.99 10 N m C )(2.00 10 C) 3.67 10 V/m.
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This note was uploaded on 06/03/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.

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ch24-p066 - 66. Since the charge distribution is...

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