CH23_2

# CH23_2 - Q2323 Linear charge density A is charge per unit...

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Unformatted text preview: Q2323 Linear charge density, A, is charge per unit length. It is used when trying to determine the electric field created by a charged rod. Surface charge density, 0-, is charge per unit area. It is used when determining the electric field above a charged sheet or disk. Volume charge density, p, is charge per unit volume. It is used when determining the electric field due to a uniformly charged sphere made of insulating material. Q2328 In special orientations the force between two dipoles can be zero or a force of repulsion. In general each dipole will exert a torque on the other, tending to align its axis with the ﬁeld created by the ﬁrst dipole. After this alignment, each dipole exerts a force of attraction on the other. P2330 5:22:88 1 8.8th2 E=2 8.99x109 7.90x104 1*——«—x~_h =4.46><108[1———x——] d )[ 11363+(0850)2 dx2+0.123 (a) At 3: = 0.050 0 m, E = 3.83 x 108 N/C = 383 MN/C (b) At x = 0.100 m, E = 3.24x 10“ N/C = (c) At x = 0.500 m, E = 8.07x 10" N/C = 80.7 MN/C (d) At x = 2.00 m, E = 6.68 x 108 N/C = 6.68 MN/C P2333 Due to symmetry E3 2 [dB]; 2 0 , and EI : IdE sing : Ice! dquzng y ' l" where dq=ﬂds=£rd6, “ 3‘5 x k ,1 \ so that, Ex 2 ke’l [sin .919: “’3” [—cos m1 = 2m 1' U r 0 r where A 2% and r = E. FIG. 112333 If 2 8.99x'109 Nurmzfc2 7.50x'10'6 C :r Thus, E, = 2h?” = ’ )2 ) . L“ (0.140 m) Solving, Ex = 2.16 x 10? N/C . Since the rod has a negative charge, E = (—2.16 x 10F N/C = 4.1.63 MN/C . q_1—6_1 qZ—IB— 3 0°) ql IS negahve, qz is positive P2340 (a) P2342 F=qE=ma az-‘E m -3151 m _ (1.602 x10_19)(520){48.0 x104) 1318ng I}? — —9.11-;<—10—_:-ﬁ‘—-“—'-= >( 106 I'll/S in a direction opposite to the ﬁeld [1.602 x 10—19)(520)(43.0 x 10—9) proton: Up — —-——1—6m‘_2?— = X 103 W5 in the same direction as the ﬁeld 0f =vi+at vf 1.602 x 10—19 6.00 x105 . P23.44 (a) lal = E = (—-———)[—) = 5.76 x 1013 m/s so a = —5.76 x 10131 mjs2 m (1.67 x 10‘”) (b) vf=vi+23(xf—x,-) 0 2 a} + 2(—5.76 x 1013 )(0070 0) v,- = 2.84 x106; m/s (c) vf=vi+at 0 = 2.84 x 106 + (6.76 x1013)t t: 4.93 x104 5 _ 3 A P23.49 v:— _ 9.55 x 10 m/s E: mo i) MC lfl—r‘ix £5 _ (1.60 x 10-19)(7zo) V 8 = __ w _....____.__ 1 _ 1 1“ 2 . (a) ay m (1.6? x 10%,) 6 90 x l] m/S Proton I Beam 3' 2 . . 2 R = W = 1.27 x 10—3 m 50 that PIC” P23_49 ﬂ 3" 3 2 - (9.55 x 10 ) 311326 3 ——m—ﬁ— = 1.27 x 10‘ 6.90 x10 5611219: 0.961 9 = 90.0°—9= R R (b) r=g=0fmsa If6=36.9°,t=. Ift9=53.1°,t=. ...
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## This note was uploaded on 09/08/2011 for the course PHYS 142 taught by Professor Staff during the Summer '08 term at Maryland.

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CH23_2 - Q2323 Linear charge density A is charge per unit...

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