Page1 / 24

Ch23F - CHAPTER 23 The Electric Field II: Continuous Charge...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon

Ch23F - CHAPTER 23 The Electric Field II: Continuous Charge...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
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 = L ( b ), ( c ), ( d ) E x ( x 0 ) = kQ /[ x 0 ( x 0 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 = +3 C/m 2 and that of the right plane is = –3 C/m 2 . Draw the electric field lines for each case. ( a ) E = 4 π k = 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- 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 ).
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
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 r 2 / x 2 ; this is not a good approximation since x = 2 r is not much greater than r . E x = 2.03 × 10 5 N/C E x = 2.54 N/C E x = 2.54 × 10 4 N/C 5* For the disk charge of Problem 4, calculate exactly the electric field on the axis at distances of ( a ) 0.04 cm and ( 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/m 2 ( b ) Proceed as in ( a ) E x = 2.00 × 10 5 N/C E x = 2.54 N/C For x = 0.04 cm, the exact value of E x is only 1.5% smaller than the approximate value obtained in the preceding problem. For x = 5 m, the exact and approximate values agree within less than 1%.
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.
Ask a homework question - tutors are online