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A copper penny has a mass of 3.0 g. A total of 4.0 × 1012 electrons are transferred from one neutral penny to another. If the electrostatic force of attraction between the pennies is equal to the weight of a penny, what is the separation between them?

Respuesta :

Answer:

F = K Q1 Q2 / R^2 = m g

Q1 = - Q2 = (4E12 * 1.60E-19) = 6.4E-7   coulombs

Q^2 = 4.10E-13

R^2 = 9.0E9 * 4.1E-13 / (.003 * 9.80) = 36.9E-4 / 2.94E-2

R^2 = .126

R = .35 m

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