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2 years ago

7

An oil droplet is levitated and held at rest in a region where the electric field is 1.4x104 N/C directed vertically downwards.

If the droplet has 3 excess electrons attached, what is its mass

1 answer:

Ratling [72]2 years ago

8 0

W = m g weight of drop

W = q E

m = q E / g = = 3 e E / g

m = 3 * 1.6E-19 * 1.4E4 / 9.8 = 6.96E-16 kg

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Answer:

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Explanation:

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$(950)^2 = 0 + 2 \times a \times 0.75\\a = 601666.67 m/s^2\\a/g = 688858.70/9.8 = 70291.70$

(b)By $v = u + at$

$950 = 0 + 601666.67 \times t\\t = 1.58 x 10^-3 sec = 1.58 ms$

(c)By $v^2 = u^2 - 2gh$

$0 = (950)^2 - 2 \times9.8 \times H\\H = 46045.92 m$

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