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 .