Electric Charge and Electric Field
21.52.
IDENTIFY: For a long straight wire, E 
SET UP:
1
2ϵ0
EXECUTE: r 
21.53.

2ϵ0r
21-25
.
 180 1010 N  m2/C2.
15 1010 C/m
 108 m
2ϵ0 (250 N/C)
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.
Q  0125  109 C, a  0025 m and x  0400 m.
→ 1
Qx
EXECUTE: (a) E 
iˆ  (70 N/C) iˆ.
2
4ϵ 0 (x  a2 )3/2
→
→
→
(b) Fon ring  Fon q  qE  (250  10 6C)(70 N/C) i ˆ (175  10 5 N) iˆ
SET UP:
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
2ϵ0 x (x/a) 2  1

 1 
 2
.

 4ϵ 0  x (x/a) 2 1
(180 109 N  m2/C2 )(175 109 C/m)
 3.03  10 4 N/C, directed upward.
2
 600 cm 
(00600 m) 
 1
 425 cm 
(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
4ϵ0 (x2  a2 )3/ 2
. We first find
the radius of the circle using 2 r  l.
EXECUTE: Solving for r gives r  l/2  (8.50 cm)/2  1.353 cm.
The charge on this circle is Q  l  (175 nC/m)(0.0850 m)  14.88 nC.
The electric field is
E
1
Qx
4ϵ 0 (x  a )
2
2 3/2

(900 109 N  m2/C2 )(1488 1029 C/m)(00600 m)
2 3/2
(00600 m) 2  (001353 m) 


E = 3.45 104 N/C, upward.
21.55.
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

 
1
.
1
charge, Ex 
2ϵ 0 
2

(R/x)
1


9
EXECUTE: The surface charge density is   Q  Q  650  10 C
A r 2  (00125 m)2
 1.324  C/m 2.
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