University Physics 13th Edition Solution Manual.pdf

May 8, 2017 | Author: Mohammed Rashwan | Category: N/A
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Electric Charge and Electric Field

21.52.

IDENTIFY: For a long straight wire, E = SET UP:

.

1 = 1.80 × 1010 N ⋅ m 2 /C2 . 2π !0 1.5 × 10−10 C/m = 1.08 m 2π !0 (2.50 N/C)

EXECUTE: r =

21.53.

λ 2π !0r

21-25

EVALUATE: For a point charge, E is proportional to 1/r 2 . For a long straight line of charge, E is proportional to 1/r. IDENTIFY: For a ring of charge, the electric field is given by Eq. (21.8). F = qE . In part (b) use

Newton’s third law to relate the force on the ring to the force exerted by the ring. SET UP: Q = 0.125 × 10−9 C, a = 0.025 m and x = 0.400 m. EXECUTE: (a) E =

1

Qx

4π !0 ( x + a 2 )3/2 2

iˆ = (7.0 N/C) iˆ.

(b) Fon ring = − Fon q = − qE = − (−2.50 × 10−6 C)(7.0 N/C) iˆ = (1.75 × 10−5 N) iˆ 21.54.

EVALUATE: Charges q and Q have opposite sign, so the force that q exerts on the ring is attractive. (a) IDENTIFY: The field is caused by a finite uniformly charged wire. SET UP: The field for such a wire a distance x from its midpoint is

E= EXECUTE: E =

1 λ 1 λ =2 . 2π !0 x ( x/a) 2 + 1 4π !0 x ( x/a) 2 + 1

(18.0 × 109 N ⋅ m 2 /C2 )(175 × 10−9 C/m) 2

6.00 cm 4.25 cm

(0.0600 m)

= 3.03 × 104 N/C, directed upward.

+1

(b) IDENTIFY: The field is caused by a uniformly charged circular wire. SET UP: The field for such a wire a distance x from its midpoint is E =

1 Qx . We first find 4π !0 ( x 2 + a 2 )3/2

the radius of the circle using 2π r = l. EXECUTE: Solving for r gives r = l/2π = (8.50 cm)/2π = 1.353 cm. The charge on this circle is Q = λl = (175 nC/m)(0.0850 m) = 14.88 nC. The electric field is E=

21.55.

1

Qx

4π ! 0 ( x + a ) 2

2 3/2

=

(9.00 × 109 N ⋅ m 2 /C2 )(14.88 × 1029 C/m)(0.0600 m) (0.0600 m)2 + (0.01353 m)2

3/ 2

E = 3.45 × 104 N/C, upward. EVALUATE: In both cases, the fields are of the same order of magnitude, but the values are different because the charge has been bent into different shapes. IDENTIFY: We must use the appropriate electric field formula: a uniform disk in (a), a ring in (b) because all the charge is along the rim of the disk, and a point-charge in (c). (a) SET UP: First find the surface charge density (Q/A), then use the formula for the field due to a disk of charge, E x =

σ 1 1− . 2!0 ( R/x )2 + 1

EXECUTE: The surface charge density is σ =

Q Q 6.50 × 10−9 C = 1.324 × C/m 2 . = 2= A πr π (0.0125 m) 2

© Copyright 2012 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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