22.
(a) Eq. 22-1 gives
F =
8.99
× 10
9
N
·m
2
/C
2
1.00
× 10
−16
C
2
(1.00
× 10
−2
m)
2
= 8.99
× 10
−19
N .
(b) If n is the number of excess electrons (of charge
−e each) on each drop then
n =
−
q
e
=
−
−1.00 × 10
−16
C
1.60
× 10
−19
C
= 625 .
22.
(a) Eq. 22-1 gives
F =
8.99
× 10
9
N
·m
2
/C
2
1.00
× 10
−16
C
2
(1.00
× 10
−2
m)
2
= 8.99
× 10
−19
N .
(b) If n is the number of excess electrons (of charge
−e each) on each drop then
n =
−
q
e
=
−
−1.00 × 10
−16
C
1.60
× 10
−19
C
= 625 .