CHAPTER23The Electric Field II: Continuous Charge Distributions1*∙A uniform line charge of linear charge density λ= 3.5 nC/m extends from x = 0 to x = 5 m. (a) What is thetotal charge? Find the electric field on the x axis at (b) x = 6 m, (c) x = 9 m, and (d) x = 250 m. (e) Find the fieldat x = 250 m, using the approximation that the charge is a point charge at the origin, and compare your resultwith that for the exact calculation in part (d).(a)Q= L(b), (c), (d)Ex(x0) = kQ/[x0(x0– L)], Equ. 23-5(e)Ex≅kQ/x2Q= (3.5×10–9×5) C = 17.5 nCEx(6) = 26.2 N/C; Ex(9) = 4.37 N/C; Ex(250) = 2.57×10–3N/CEx(250) = 2.52×10–3N/C, within 2% of (d)2∙Two infinite vertical planes of charge are parallel to each other and are separated by a distance d = 4 m.Find the electric field to the left of the planes, to the right of the planes, and between the planes (a) when eachplane has a uniform surface charge density σ= +3 µC/m2and (b) when the left plane has a uniform surfacecharge density = +3 C/m2and that of the right plane is = –3 C/m2. Draw the electric field lines for eachcase.(a)E= 4πk= 3.39×105N/CThe field pattern is shown in the adjacent figure.The field between the plates is zero.(b) Again, E= 3.39×105N/C.The field pattern is shown in the adjacent figure.The field is confined to the region between thetwo plates and is zero elsewhere.3∙A 2.75-C charge is uniformly distributed on a ring of radius 8.5 cm. Find the electric field on the axis at(a) 1.2 cm, (b) 3.6 cm, and (c) 4.0 m from the center of the ring. (d) Find the field at 4.0 m using theapproximation that the ring is a point charge at the origin, and compare your results with that for part (c).
Chapter 23 The Electric Field II: Continuous Charge Distributions(a) Use Equ. 23-10(b), (c) Proceed as in (a)(d)Ex~ kQ/x2Ex=N/C10_4.69=N/C)(0.085)+((0.012)0.012_10_2.75_108.99_53/22269•Ex(0.036) = 1.13×106N/C; Ex(4) = 1.54×103N/CEx= 1.55×103N/C; this is slightly greater than (c)because the point charge is nearer x= 4 m than the ring.4∙A disk of radius 2.5 cm carries a uniform surface charge density of 3.6 µC/m2. Using reasonableapproximations, find the electric field on the axis at distances of (a) 0.01 cm, (b) 0.04 cm, (c) 5 m, and (d) 5cm.For x<< r, the disk appears like an infinite plane. For x>> r, the ring charge may be approximated by a point charge.(a), (b) Use Equ. 23-12a(c)Ex= kQ/x2= kπr2σ/x2(d)Ex= kr2/x2; this is not a good approximationsince x= 2ris not much greater than r.Ex= 2.03×105N/CEx= 2.54 N/CEx= 2.54×104N/C5*∙For the disk charge of Problem 4, calculate exactly the electric field on the axis at distances of (a) 0.04 cmand (b) 5 m, and compare your results with those for parts (b) and (c) of Problem 4.(a) Use Equ. 23-11; r= 2.5 cm, = 3.6 C/m2(b) Proceed as in (a)Ex= 2.00×105N/CEx= 2.54 N/CFor x= 0.04 cm, the exact value of Exis only 1.5% smaller than the approximate value obtained in thepreceding problem. For x= 5 m, the exact and approximate values agree within less than 1%.
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