A piece of dust, with mass 0.1 g, is in between two charged parallel plates. Ini

Question

A piece of dust, with mass 0.1 g, is in between two charged parallel plates. Initially the electric field between the plates is 2000 N/C, and the dust particle is at rest, and suspended in the air between the two plates. (a) What is the magnitude of the charge on the dust? A laser is used to shine light on the particle which results in removing electrons from the dust. After the light is applied, the particle begins to fall vertically downward, and it is found that the electric field must be increased to 2500 N/C in order to again have the particle be at rest. (b) Neglecting any change is mass, what is the new charge on the piece of dust? (c) Assuming electrons were either added or removed, determine how many electrons were added or removed from the dust particle by the light? Follow-up: By what percentage would this change the mass of the dust particle, and was it OK to neglect this change?

Explanation / Answer

(A) magnitude of electrical force = magnitude of gravittaional force

q E = m g

q = (0.1 x 10^-9 kg) (9.8)/ (2000)

q = 4.9 x 10^-13 C

(B) q = (0.1 x 10^-9) (9.8) / 2500

q = 3.92 x 10^-13 C

(C) delta(q) = n e

(4.9 - 3.92) x 10^-13 = (n)(1.6 x 10^-19)

n = 612500 electrons

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