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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

Answer :

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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