Chapter24solns

Chapter24solns - 1. What is the net charge shown in Fig....

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Unformatted text preview: 1. What is the net charge shown in Fig. 24-39? The magnitude of the middle charge is 3 ac. Solution The number of lines of force emanating from (or terminating on) the positive (or negative) charges is the same (14 in Fig. 24—39), so the middle charge is I—3 ac and the outer ones are +3 ,uC. The net charge shown is therefore 3 + 3 — 3 = 3 ac. This is reflected by the fact that 14 lines emerge from the boundary of the figure. FIGURE 24-39 Problem 1 Solution. 3. Two charges +q and a charge —q are at the vertices of an equilateral triangle. Sketch some field lines for this charge distribution. Solution (The sketch shown follows the text‘s convention of eight lines of force per charge magnitude q.) Problem 3 Solution. 5. A flat surface with area 2.0 in2 is in a uniform electric field of 850 NEC. What is the electric flux through the surface when it is (a) at right angles to the field, (b) at 45” to the field, and (c) parallel to the field? Solution (3.) When the surface is perpendicular to the field, its normal is either parallel or anti~parallel to E. Then Equation 24—1 gives ‘3 = E-A : EA cos(0° or 180°) 2-: :l:(35l] N/C)[2 m2) 2: i170 hN-m2/C. (b) (I) E-A = EA cos(45c or 135°) emu kN-mQ/CM (0.866) 2 i120 kN-n1?,I’C. (c) <1: = EA cos 90° = 0. i’. A flat surface with area 0.14 In2 lies in the 3-3; plane, in a uniform electric field given by E = 5.11 + 2.13 + 3.51} M30. Find the flux through this surface. Solution The surface can be represented by a vector area A = (0.14 m2)[il}). (Since the surface is Open, we have a choice of normal to the x—y plane.) Then <1: : E-A = :hE-fcxwii n12) = $1340.14 m2) = has kN/C)(o.14 m2) e490 xhfijc. (Only the 2 component of the field contributes to the flux through the rug plane.) 11. What is the electric flux through each closed surface shown in Fig. 24-43? FIGURE 2&43 Problem 11. Solution From Gauss’s law, ‘13 2 qenclosed/Eo. For the surfaces shown, this is (a) (q — 290/50 2 —q/sg, (b) —2q/eo, (c) and (d) l]. 13. A 2,311.0 charge is at the center of a cube 7.5 cm on each side. What is the electric flux through one face of the cube? Hint: Think about symmetry, and don’t do an integral. Solution The symmetry of the situation guarantees that the flux through one face is % the flux through the whole cubical surface, so Guam = % fcubeE - (LA 2 cameraman (2.6 hC)/6(s.85x10-12 Cg/N-rng) = 49.0 kN-mEg’C. 1?. The electric field at the surface of a unifornilv charged sphere of radius 5.0 cm is 90 kNg'C. What would he the field strength 10 cm from the surface? The electric field due to a uniformly charged sphere is like the field of a point charge for points outside the sphere, i.e., E(r) ~ Uri for r 2 R. Thus, at 10 cm from the surface, it" : = 15 cm and Ef15 cm) -: l5;’l§)2Ex(5 cm) :: (90 kl\'/C),x'9 -:: 10 kNg'Cl. 25. A spherical shell 30 cm in diameter carries a total charge 85 ,uC distributed uniformly ever its surface. A 1.0-uC point charge is located at the center of the shell. What is the electric field strength (a) 5.0 cm from the center and 45cm from the center? (c) How would your answers change if the charge on the shell were doubled? Solution (a) The field due to the shell is zero inside, so at r: 5 cm, the field is due to the point charge only. Thus, E = heir/7'2 = (9x109 N-m2/C2)(1 ,uC)f'/(U.05 mp: (3.60x105 i\’/C)i". (b) Outside the shell, its field isle, that of a point charge, so at r = 45 cm, E = k(q+ one»? = (9x109 N-m2/Cz)(86 ,uC)f'/(0.45 m)2 = (3.82x105 N/C)f. (c) If the charge on the shell mere doubled, the field inside would be unaffected, whiletlie field outside would approximately double, E : k(1.0 so + 2x85 MC)/(45 m)2 = 7.60 MN/C. 26. The thick, spherical shell of inner radius a. and outer radius .5 shown in Fig. 2445 carries a uniform volume charge density p. Find an expression for the electric field strength in the region tr < r < b, and show that your result is consistent with Equation 244’ when a = 0. Solution Use the result of Gauss’s law applied to a spherically symmetric distribution, E = qenclosed/4rr50r2. For a < r < b in a spherical shell with charge densityr p, genclosed : §W(T3 _ a3)pi 50 E = P(?‘3 _' G3);l350r2= (p,"3ao)(r -— (ta/T2). If a ——> 0, Equation 24-? for a uniformly charged spherical volume is recovered. Gaussian surface FIGURE 24-45 Problem 26 Solution. 31. An infinitely long rod of radius R carries a uniform volume charge density ,0. Show that the electric field strengths outside and inside the rod are given, respectively, by E =: pR2/250r and E = pr/2sg, where r is the distance from the rod 3X15. Solution The charge distribution has line symmetry (as in Problem 29) so the flux through a coaxial cylindrical surface of radius in (Equation 24—8) equals qcnclflsed/so, from Gauss’s law. For T > R (outside the rod), genclosed = {’31-sz3 hence Eout : WR2£f27W£80 : pita/2501'. For 1' < R (inside the rod), qenclosed = piTT2g, hence Eu 2 pirr2£f2irr€sg == {JTfr’IQ'L-Tg. (The field direction is radially away from the symmetry axis if p > D, and radially inward if p < O.) 35. If you “painted” positive charge on the flour, what: surface charge density would be necessary in order" to suspend a 15-pC, 5.0-g particle above the lime?- Solution A positive surface charge density or, on the floor, would produce an approximately uniform electric field upward of E 2 0/250, at points near the floor andiiet' near an edge. The field needed to balance the weight I of a particle, of mass m. and charge 9, is given by my = qE, therefore 0 : 250E = 250mgg'q = 2(835}: 10-12 C2/N-m2)(5x10_3 kg)(9.8 m/sglfllhx in—6 C) = 57.8 nC/mz. 39. A nonconducting square plate 75 cm on a side carries a uniform surface charge density. The electric field strength 1 cm from the plate, not near an edge, is ‘15 kN/C. What is the approximate field strength 15 m from the plate? Solution The electric field strength close to the plate (1 cm << 75 cm) has approximate plane symmetry (E : : 0,5250), so the charge on the plate is g = 011 = 2591314 = 2(8.85x10"'12 C2/’N-n12}[45 kN;’C)x (0.75 m}2 :2: 448 nC. Very far from the plate (15 In 3) 0.75 m), the field strength is like that from a point charge, E = kqfr‘? 2 (9X109 Ema/(72V {448 nC)(15 Inf—Q = 17.9 N30. 45. A net charge of 5.0 ,uC is applied on one side of a solid metal sphere 2.0 cm in diameter. After electrostatic equilibrium is reached, what are (a) the volume charge density inside the sphere and (b) the surface charge density on the sphere:J Assume there are no other charges or conductors nearby. (c) Which of your answers depends on this assumption, and why? Solution (a) The electric field within a conducting medium, in electrostatic equilibrium, is zero. Therefore, Gauss’s law implies that the net charge contained in any clcsed surface, lying within the metal, is zero. (b) If the volume charge density is zero within the metal, all of the net charge must reside on the surface of the sphere If the sphere is electrically isolated, the charge will be uniformly distributed (i.e., spherically symmetric}, so or = Q/xirrR2 = (5 nC)/Llrr(1 cm}2 2 3.98x 10—3 Cfml (c) Spherical symmetry for or depends on the proximity of other charges and conductors. iii'. A 25U—nC point charge is placed at the center of an uncharged spherical conducting shell 20 cm in radius. (a) What is the surface charge density on the outer surface of the shell? (b) What is the electric field strength at the shell’s outer surface? Solution is) There is a non-zero field outside the shell. because the net charge within is not zero. Therefore, there is a surface charge density 0' :- eoE on the outer surface of the shell, which is uniform, if we ignore the possible presence of other charges and conducting surfaces outside the shell. Gauss‘s law (with reasoning similar to Example 24—7) requires that the charge on the Shell’s outer surface is equal to the point charge within, 50 o = (3/41ng = 250 nC/4?r(0.20 m)"2 s 497 Dogma. - {in} Then the field strength at the outer surface is 5‘: 0/50 = 56.2 kN/C. Apoint charge q is at- the center of a spherical shellef radius R carrying charge 2:} spread Solution - lR <. R. - - -' for T 2: As explained in Example 24 5; (a) g 2 _ qencmscd : q and E = fig/CERF ;: 4M“? . gulf}; {M2 :2: 2R > Rs qenclosed :: q + 2‘; and E :I: Skq/I‘LR} " 3 {cg/4R2. ...
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Chapter24solns - 1. What is the net charge shown in Fig....

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