74. (a) Let E = σ/2ε0 = 3 × 106 N/C. With σ = |q|/A, this leads to 2
c
hc
h
2.5 × 10−2 m 3.0 × 106 N C R2 E 2 2 = 10 q = πR σ = 2 πε 0 R E = = . × 10−7 C. 9 N ⋅m 2 2k 2 8.99 × 10 C2
d
i
(b) Setting up a simple proportionality (with the areas), the number of atoms is estimated to be N=
c
h
π 2.5 × 10−2 m
2
0.015 × 10−18 m2
= 13 . × 1017 .
(c) Therefore, the fraction is q 1.0 ×10−7 C = ≈ 5.0 × 10−6. 17 −19 Ne (1.3 × 10 ) (1.6 ×10 C )