University Physics 13th Edition Solution Manual.pdf

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

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