Chapter25solns

Chapter25solns - Chapter 25 Solutions ‘:. How much work...

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Unformatted text preview: Chapter 25 Solutions ‘:. How much work does it take to move a 50-pC charge against a 12-V potential difference? Solution I The potential difference and the work per unit charge, dune by an external agent, are equal in magnitude, so ll’quV = (50 ,uC}(12 V) --= 600 ,uJ. (Note: Since ally magnitudes are needed in this problem, we outfitted the subscripts A and B.) 12. Electrons in a TV tube are accelerated from rest through a 25-kV potential difference. With what speed do they hit the TV screen? Solution The work done on an electron equals the change in its} kinetic energy, W = e AV 2 é’m’n2 (if it starts from l rest). Thus, r——'_——-———— 2(1.6><10'19 C)(25x103 V) ==,/ Ar! = _________ 1'. 2e Um (gllxlmm kg) = 9.37x10? m/s. 13. A l2-V car battery stores 2.8 MJ of energy. How much charge can move between the battery terminals before it is totally discharged? Assume the potential dilferenoe remains at 12 V, an assumption that is not realistic. I Solution A charge q, moving through a potential difference all. is equivalent to electrostatic potential energy AU : q AV. stored in the battery. Thus, 9 = 2.8 MRI/12V: 2. 33 x 105 C. 17. A 5.0-g object carries a net charge of 3.8 ,uC. It acquires a speed a when accelerated from rest through a potential difference V. A 2.0-g object acquires twice the speed under the same circumstances. What is its charge? Solution The speed acquired by a. charge q, starting from rest at point A and moving through a potential difference of V, is given by %mc§ = gal, — V3) = qV, or 1-‘3 : ,ZQl’lq/m). (This is the work—energy theorem for the electric force. A positive charge is accelerated in the direction of decreasing potential.) If the second object acquires twice the speed of the first object, moving through the same potential difference, it must have four times the charge to mass ratio, qZ m. Thus. Q2 2 4(q1fm1lm2 = 4(3.8 nC)(2g/5g) = 6.08 pic. 18. An electric field is given by E = E03, where E0 is a constant. Find the potential as a function of position. taking V = 0 at y = 0. Solution Since this is a uniform field, Equations 25—2a or b give Vly) * Viol = Vth = —E-«'3 = *{onl‘lr - 0) = ‘Efl'y. 31 Three equal charges q form an equilateral triangle of side c. Find the potential at the center of the triangle. Solution l The center is equidistant from each vertex, and r = alEcos30° = a/ Each charge contributes equally lathe potential, so V = 3kq/T 7— Sofikq/a. ‘ Problem 30 Solution. 35. A hollow, spherical conducting shell of inner radius h and outer radius c surrounds, and is concentnc with, a solid conducting sphere of radius a, as shown in Fig. 25-39. The sphere carries a net charge ——Q and the shell carries a net charge +362. Both conductors are in electrostatic equilibrium. FIGURE 25-39 Problem 35. Find an expression for the potential difference from infinity to the surface of the sphere. Solution The electric field between the solid sphere and the sheil is like that due to a point charge —-Q located at their common center (the origin), so the potential difference between the sphere and the shell is Vs h — 14,,“ = —kQ(o‘1 — if"). The electric field outsirde the shell is like that due to a point charge 262 at the origin (the total charge is 3Q — ), so the potential difference between the shell and infinity is ion,“ — V00 = 2ch‘1. (See Examples 24—3 and 25—3.) The entire shell is at one potential, as is the entire sphere, because each is a conductor in electrostatic equilibrium. Therefore the potential difference between the sphere and infinity is Vsph — We = sph - Vshell + Vsheu -- V = “2964 “t 5‘1 — (1"1}. I 37. thin ring of radius R carries a charge 3Q distributed uniformly over three-fourths of its circumference, and —Q over the rest. What is the potential at the center of the ring? Solution The result in Example 25-6 did not depend on the ring being uniformly charged. For a point on the axis of the ring, the geometrical factors are the same, and Ling do = Qtot for any arbitrary charge distribution, so l’ = kgtottcz + e2)"1/2 still holds. Thus, at the center {I 2-: 0) 'of a ring of total charge th = 362 — Q = 2Q and radius a = R, the potential is V :- 2kQ/R. , 4?. The electric potential in a region of space is given by i" = 233,: — 323 + 53:2, with V in volts and the coordinates in meters. If point P is at :r 2: 1 In, y: 1 in, z r: 1 in, find (a) the potential at P and (b) the :r, y, and 3 components of the electric field at P. Solution Direct substitution gives V(P) = 2(1){1) — 3{1}(1) + 5(1)2 2 4 V. {'0} Use of Equation 25—10 gives l?I :— —6l"f5.r = T —2y + 3z, Ey = -3V;'5y = —2x — 103;, and E; = ——6V;’6z 2 3st. At P(I = y = z = 1), E1 =1Wm. E3, :-= ——12 V/m, and l3z = 3 V/ni. . The electric potential in a region is given by V : —ir’g(r,-’R), where l?) and R are constants, r is the radial distance from the origin, and where the zero of potential is taken at r = 0. Find the magnitude and direction of the electric fieid in this region. Solution Since l2" 2 -(V0fR)T depends only on i", the field is spherically symmetric and its direction is radiai. From Equation 25—10 (which applies unmodified for the radial coordinate), E.— =-— ——dV/dr : = l/b/R, and E = (VDIR)? (f- is a unit vector radiain outward). 53. The spark plug in an automobile engine has a center electrode made from wire 2.0 min in diameter. The electrode is worn to a hemispherical shape, so it behaves approximately like a charged sphere. What is the minimum potential on this electrode that will ensure the plug sparks in air? Neglect the presence of the second electrode. Solution If we can treat the field from the central electrode as that from an isolated sphere, then E = kq/R2 and V = Esq/R, so that V 2 RE. Dielectric breakdown in air would occur for potentials exceeding V = (1 mm)x (3X106 V/m) = 3 kV. 53. Two small metal spheres are located 2.0 m apart. One has radius 0.50 cm and carries 0.20 iiC. The other has radius 1.0 cm and carries 0.080 nC. (a) What is the potential difference between the spheres? (b) If they were connected by a thin wire, how much change would move along it, and in which direction? Solution {a} Since the spheres are approximately isolated, their potentials are V, = kql/Rl = (9x109 V-m/C)>< Bill)"r C-)/(5><10_3 rn) = 3.6x10" V, and V2 :-5 (9x109 V-m/‘C)(8x10—8 C)/(1o-2 HI) = 0.72x10 V. The difference is V1 r V; = 2.88x105 V. (b) The total charge on the spheres is 0.28 ,uC = qi +q’2, where primes refer to the spheres after they are connected. {We assume that the wire has a negligible effect, other than bringing both spheres to the same potential.) Since l"; = kq’l/Rl = V’ = kq'g/Rg, ngi - qu2 =—- D, oi2q’1 v o; = 0. Solving for qi, We find g“1 = £0.23 p0) = 0.0933 JuC, therefore a charge of Q1 — q’l = 0.107 ,uC was transferred from the first sphere to the second. ...
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This note was uploaded on 08/23/2008 for the course PHYS 2b taught by Professor Schuller during the Summer '08 term at UCSD.

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Chapter25solns - Chapter 25 Solutions ‘:. How much work...

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