21.4: Mass of gold = 17.7 g and the atomic weight of gold is 197 g mol. So the number
of atoms = N A " mol = (6.02 " 10 ) "
23
a)
(
17.7 g
197 g mol
) = 5.41" 10
22
.
np = 79 " 5.41 " 1022 = 4.27 " 1024
q = np " 1.60 " 10#19 C = 6.83 " 105 C
b) ne = np = 4.27 " 1024.
r r r
21.20: F = F1 + F2 and F = F1 " F2 since they are acting in opposite directions at
x = 0 so,
& 4.00 $ 10 %9 C 5.00 $ 10 %9 C )
1
%6
F=
6.00 $ 10 %9 C (
%
N to the right.
2
2 + = 2.4 $ 10
4"#0
(0.200
m)
(0.300
m)
'
*
21.22: a)
b)
F x = "2
!
1
qQ
1
"2qQx
, F =0
2
2 cos% =
4#$ 0 (a + x )
4#$ 0 (a 2 + x2 )3 / 2 y
c) At x = 0, F = 0.
d)
21.70: Examining the forces: " F x = T sin# $ F e = 0 and " F y = T cos# $ mg= 0.
So
mgsin "
cos "
= Fe =
kq2
d
2
. But tan" # 2dL $ d =
3
2kq 2 L
mg
$d=
(
)
q 2L
2 %& 0 mg
1 3
.