電學一問, 多多指教

1. By E = V / d
E = (2 X 103) / 0.05 = 4.0 X 104 NC-1
Since the oil drop is at rest, by Newton’s 2ndlaw of motion
mg – qE = 0
mg = qE
10m = (1.6 X 10-19 X 10-3)(4.0 X 104)
m = 1.6 X 10-19 kg (For 1.6 X 10-19 mC) (Just take the magnitude ofthe charge)
mg = qE
10m = (1.6 X 10-19)(4.0 X 104)
m = 1.6 X 10-16 kg (For 1.6 X 1019 C)
I guess maybe the second answer is the correct answer.Since 1.6 X 10-19 C is the foundamental charge in the nature. Thereis no such 1.6 X 10-19 mC. However, I just did it for yourreference.

Using Shell’s theorem, you can solve out that inside allconductors, the electric field is zero.
Here, I would like to explain this fact without usingShell’s theorem.
If there is an electric field inside the conductor, therewill be a current flow inside the conductor and hence it is not inelectrostatics. Since charges are free to flow inside a conductor. This isundesirable. Hence we can conclude that inside all conductors, the electricfield is zero.

2008-03-15 11:42:55 補充：
Sorry, 答案唔記得multiple by 4
mg = qE

10m = (1.6 X 10^-19)(4.0 X 10^4)

m = 6.4 X 10^-16 kg (For 1.6 X 10^19 C)
我都唔太理解點解要計mg，因為應該要係vertical先會計mg。如果啲field係magnetic field就係horizontal了。