Asked by Happiness delight
The earth has an electric field of about 150 N/c pointed downward. A 1.00 (micrometer) radius of water droplet is suspended in a calm air. Find the
1) mass of the water droplet
2) charge on the water droplet
3) number of access electrons in the water droplet.
1) mass of the water droplet
2) charge on the water droplet
3) number of access electrons in the water droplet.
Answers
Answered by
Damon
volume of drop = V = (4/3) pi r^3
mass of drop = m = 1000 * V
gravity down = F = m g
force up = Q E = F
number of excess electrons = Q /e
mass of drop = m = 1000 * V
gravity down = F = m g
force up = Q E = F
number of excess electrons = Q /e
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