Chapter 23 Homework

Chapter 23 Homework - Q23.6 The electric field due to the...

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Unformatted text preview: Q23.6 The electric field due to the charged rod induces charges on near and far sides of the sphere. The attractive Coulomb force of the rod on the dissimilar charge on the close side of the sphere is larger than the repulsrve Coulomb force of the rod on the like charge on the far side of the sphere. The result 15 a net attraction of the sphere to the rod. When the sphere touches the rod, charge is conducted between the rod and the sphere, leavin both the rod and the s h . _ results in a repulsive Coulomb force. g p ere like-Charged" “‘15 Q23.11 So the electric field created by the test charge does not di measure, by moving the charges that create it. stort the electric field you are trying to Q2315 If a charge distribution is small compared to the distance of a field point from it, the charge distribution can be modeled as a single particle with charge equal to the net charge of the distribution. Further, if a charge distribution is spherically symmetric, it will create a field at exterior points just as if all of its charge were a point charge at its center. Q23.16 The direction of the electric field is the direction in which a positive test charge would feel a force when placed in the field. A charge will not experience two electrical forces at the same time, but the vector sum of the two. If electric field lines crossed, then a test charge placed at the point at which they cross would feel a force in two directions. Furthermore, the path that the test charge would follow if released at the point where the field lines cross would be indeterminate. “ ' ' ' 7 r ' V f ulsion. In general 23.28 In speaal orientations the force between two dipoles can be zero or a 0 rep Q 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. ‘ ’— 10.0 grams 23 atoms)( electrons) 7A . N = ———— 6.02 10 47 —— = 2.62 10 P23 2 (a) [107.87 grams/mol x mol atom —3 (b) # electrons added = 9 = = 6.25 x 1015 e 1.60x10'19 C/electron or 2.38 electrons for every 109 already present . P233 If each person has a mass of z 70 kg and is (almost) composed of water, then each person contains N E m (6.02 x 1023 w) 10 m) E 2.3 x 102’8 protons. 18 grams/mol mol molecule With an excess of 1% electrons over protons, each person has a charge q = 0.01(1.6 x 10-19 c)(2.3 x 1028) = 3.? x 107 c . (3.7 x 107)2 TN=4X1025 This force is almost enough to lift a weight equal to that of the Earth: Mg = 6x1034 kg(9.8 m/sz)=6 x 1025 N~10?" N. So F=ke ‘7qu =(9x109) ;, n. r k *P23.4 The force on one proton is F = “17qu away from the other proton. Its magnitude is r- -19 2 (8.99x109 N-m/C2)(lé6—:1—1(:)Tl§-n%] =. P23.7 P23.10 *P23.17 P7319 P2325 P2326 qlqz = (8.99x109 N-mz/c2)(7.00x10* c)(2.00x10* c) P1 = k. = 0503 N r2 (0.500 m)2 F1 9 2 2 ' _ y F2 :k M2 = (8.99x10 N.m /C )(7.00x10* C)(4.00x10 6 C) _1 01 N ®7_oo,,c r“ (0.500 In)2 . 1:2 0.500 m o 2.00 60° F}. = 0.503 cos 60.0 +1.01 cos 60.0°= 0.755 N 140+ — x Fy = 0.503 sin 60.0°—1.01 sin 60.0°= —0.436 N ‘ 4'00 “C F = (0.755 N)i — (0.436 N); =) 0.872 N at an angle of 330° ) FIG. P23.7 Let the third bead have charge Q and be located distance x from the left end of the rod. This bead will experience a net force given by k 3 A k A x (d-x) . . 3 1 x Thenetforcewfllbezerorf7=—7,ord—x=—. x (d—x) J3 This gives an equilibrium position of the third bead of x = 0.634d . The equilibrium is stable if the third bead has positive charge . The first charge creates at the origin field I“? to the right. +CQ x T 0 I I 5 x a Suppose the total field at the origin is to the right. Then q must be negative: FIG. P23.17 2 k : 2k : k;§21+(3:)72(—1)=-i221 . In the alternative, the total field at the origin is to the left: '2? a + :5: (4) = Eg—QH) . kg , 8.99 x 109 3.00 x 10'9 2 1 y (a) 151 = rllzli (-3) = (__(T.1)(Fo)_2—)(_,) = —(2.70 x 103 N/C)) 52 6.00 If x k . 8.99 109 6.00 x 10‘9 t 2 E IE1 E2 = (-i) = ( " (0.3120): )(-1) = —(5.99 x 102 me); @400 “C E = E2 + E1 = —(5.99 x 102 N/C)i — (2.70 x 103 N/C)i FIG. P23.19 (b) 1: = qE = (5.00 x 10-9 c)(—599§ — 2 7003) N/c F = (-3.00 x 10* i — 13.5 x 10* i) N = (—3.00i-13.5}) #N _ ken k,(Q/€)£ _ keg _ (3.99 x 109)(22.0 x 10*) g ‘ d(l + d) ‘ 4(2 + d) ' d(£ +d) = (0.290)(0.140 +0290) :<—d—>('E “‘36.0 cm—w E = 1.59 x 106 N/C, directed toward the rod. FIG. P2325 k E=I ‘gq ,where dq=lodx x Q k A A = The direction is — i or left for ’10 > 0 x x0 0 °° dx 1 E = kelo j—z = Ice/10(7) x0 x P2328 P2333 P2335 P23.40 P23.41 wklde—i :m_ c 100 : E=jdE= Jo “_°:T(_) =_k,,10x01£ x 3dx=—kelox01[—§Lo]= 5x00(_l) ' 0 Duetosymmetry Ey=JdEv=0,and Exzdesin0=deq512n y ' r where dq = Ms = me, e x " k 2k K so that, )5x = kel Isin0d0= “'1 (~C050)|: = *1 r 0 r where = % and r = —I:- I FIG. P2333 71' 2k q, 2(8.99x1o9 N-mz/C2)(7.50x10‘6 c)” Thus, Ex = ‘3 =-—————2——. L (0.140 m) Solving, Ex = 2.16 x 107 N/C. Since the rod has a negative charge, E = (—2.16 x 107 N/C = —21.6i MN/ . (a) The electric field at point P due to each element of length dx, is dE = 5‘“ 2 and is directed along the line joining the element to x + y point P. By symmetry, )5x = IdEx = 0 and since dq = Mx , I 15:15 =IdEy=JdEcosl9 where c056=-——y———. 0” d" 5' x2 + yz —’ FIG. P2335 1/2 . Therefore, [5: 2kg 1y J’ dx 3/2 = Zkel sm 00 I 0 (x2 + yz) y (b) For a bar of infinite length, 90 = 90° —6 1 — 11“ qz 3 (b) ql is negative, qz is positive (a) The electric field has the general appearance shown. It is zero , where (by symmetry) one can see that the three charges individually produce fields that cancel out. In addition to the center of the triangle, the electric field lines in the second figure to the right indicate three other points near the middle of each leg of the triangle where E = 0 , but they are more difficult to find mathematically. or; (b) You may need to review vector addition in Chapter Three. The electric field at point P can be found by adding the electric field The electric field from a point charge is E = ke Liz-f. r As shown in the solution figure at right, E1 = kL. all to the right and upward at 60° E2 = kg 12 to the left and upward at 60° a FIG. P23.41 E = 151+ E2 = e iz[(coseo°i +sin60°§) +(— cos60°i + sin60° = ke i,[2(sin60° a a- = 1.73k, 12} -19 _qE_1.602X10 614X1010 $2 P2343 (a) a _ —m —1.67 x 1047 m/ (b) v, = 2); + at 1.20 x 106 = (6.14x 101°): t = 1.95 x 10-5 s 1 1 (c) x, —x,. =§(v,. +vf)t x, =E(1.20x106)(1.95x10_5)= 2 — (d) K = émvz =%(1.67 x 10'27 kg)(1.20 x 106 m/s) = 1.20 x 10 ‘5 1 P23.45 The required electric field will be . Work done = AK 50, —Fd = -%mvi2 (since the final velocity = O ) which becomes eEd = K and E=—. x 0.0500 -7 P23.47 a t=—=‘=1.11 10 s: 111 ns ( ) 0 450x105 x _ qE _ (1.602 x 10-19)(9.6ox 103) M ay — 7 - (1.67x 10‘” ) =9.21x10“ m/s2 1 2, _1 n —7 2_ —3 _ yf-yi=vyit+§ayt. yf—§(9.21x10 )(IllxlO ) —5.68x10 m—W (c) v, = 4.50x105 m/s vyf =vyl-+ayt=(9.21x10")(1.11x104): 1.02x105 m/s ...
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This homework help was uploaded on 04/19/2008 for the course PH 2223 taught by Professor Seong-gonkim during the Spring '08 term at Ole Miss.

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Chapter 23 Homework - Q23.6 The electric field due to the...

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