P22_022.PDF

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Chapter 22 - 22.22
22. (a) Eq. 22-1 gives
8 . 99
×
10 9 N
·
m 2 / C 2 1 . 00
×
10 16 C 2
F =
=8 . 99
×
10 19 N .
(1 . 00
×
10 2 m) 2
(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
= 625 .
1 . 60
×
10 19 C
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