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8B-Soln03-v1.0.4 - 9 N/coul f From our answer to part(c E r...

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7. The charge distribution in this problem is spherically symmetric , so the Electric field has the form r E = Q enc ( r ) 4 "# 0 r 2 ˆ r at all points. Q enc ( r ) is the net amount of charge enclosed by an imaginary sphere of radius r , centered at the origin, r is the radial distance from the origin and ˆ r points radially outward from the origin. a) At r = 0.5 m, Q enc ( r ) = +2 C, so the E -field points radially outwards . b) | E ( r = 0.5 m)| = 7.2 × 10 10 N/coul (2 coulombs is a very large amount of charge!) c) At r = 1.5 m, we are inside the metal shell, which is a conductor . Therefore, the E - field must be zero! d) At r = 2.5 m, Q enc ( r ) = -1 C, so the E -field points radially inwards . e) | E ( r = 2.5 m)| = 1.4 ×
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Unformatted text preview: 9 N/coul f) From our answer to part (c), E ( r = 1.5 m) = 0, so Q enc ( r = 1.5 m) = 0. The only place a net charge can live on a conductor in an electrostatic situation is on the surfaces of the conductor, so Q enc = +2C + Q inner surface = 0, so Q inner surface = -2 C. g) The conductor has a net charge of -3 C, so Q outer surface = -1 C. h) 8. a) E ( r = R ) = b) E ( r = 3 R ) = " 2 #$ R c) r F = q e R ˆ x 9. a) III > I = II > IV ( E I = E II = 3/2 E III = 2 E IV ) b) IV > II > III > I ( E I = 0; E II / 2 = E III = E IV / 3) c) The electron will hit the right plate (with +2 Q of charge on it) d) " t = 3 m e dA # 2 q e Q...
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