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**Unformatted text preview: **CHAPTER 23 The Electric Field II: Continuous Charge Distributions 1* ∙ A uniform line charge of linear charge density λ = 3.5 nC/m extends from x = 0 to x = 5 m. ( a ) What is the total 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 field at x = 250 m, using the approximation that the charge is a point charge at the origin, and compare your result with that for the exact calculation in part ( d ). ( a ) Q = λ L ( b ), ( c ), ( d ) E x ( x ) = kQ /[ x ( x – L )], Equ. 23-5 ( e ) E x ≅ kQ / x 2 Q = (3.5 × 10 –9 × 5) C = 17.5 nC E x (6) = 26.2 N/C; E x (9) = 4.37 N/C; E x (250) = 2.57 × 10 –3 N/C E x (250) = 2.52 × 10 –3 N/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 each plane has a uniform surface charge density σ = +3 µ C/m 2 and ( b ) when the left plane has a uniform surface charge density σ = +3 µ C/m 2 and that of the right plane is σ = –3 µ C/m 2 . Draw the electric field lines for each case. ( a ) E = 4 π k σ = 3.39 × 10 5 N/C The field pattern is shown in the adjacent figure. The field between the plates is zero. ( b ) Again, E = 3.39 × 10 5 N/C. The field pattern is shown in the adjacent figure. The field is confined to the region between the two plates and is zero elsewhere. 3 ∙ A 2.75- µ 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 the approximation 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 ) E x ~ kQ / x 2 E x = N/C 10 _ 4.69 = N/C ) (0.085) + ((0.012) 0.012 _ 10 _2.75_ 10 8.99_ 5 3/2 2 2 6 9 • E x (0.036) = 1.13 × 10 6 N/C; E x (4) = 1.54 × 10 3 N/C E x = 1.55 × 10 3 N/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 µ C/m 2 . Using reasonable approximations, find the electric field on the axis at distances of ( a ) 0.01 cm, ( b ) 0.04 cm, ( c ) 5 m, and ( d ) 5 cm. 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. 23-12 a ( c ) E x = kQ / x 2 = k π r 2 σ / x 2 (d) E x = k π r 2 σ / x 2 ; this is not a good approximation since x = 2 r is not much greater than r ....

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