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saveliy_v [14]
3 years ago
6

A small piece of dust of mass m = 4.1 µg travels through an electric air cleaner in which the electric field is 466 N/C. The ele

ctric force on the dust particle is equal to the weight of the particle.
(a) What is the charge on the dust particle?
C

(b) If this charge is provided by an excess of electrons, how many electrons does that correspond to?
electrons
Physics
1 answer:
pantera1 [17]3 years ago
7 0

Answer:

Explanation:

mass, m = 4.1 x 10^-6 g

Electric field, E = 466 N/C

Electric force = weight

(a) Electric force = q x E = m x g

where, q be the charge on the dust particle.

q x 466 = 4.1 x 10^-6 x 9.8

q = 8.62 x 10^-8 C

(b) Number of electron  = charge / charge of one electron

n = 8.62 x 10^-8 / (1.6 x 10^-19)

n = 5.4 x 10^11

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Answer:

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From the question we are told that

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Generally according to Bernoulli's  equation

        p_1 =  [\frac{1}{2}  \rho v_2^2 + h_2 \rho g +p_2] -[\frac{1}{2} \rho * v_1^2 + h_1 \rho g ]

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        p_2 is the gauge pressure of the atmosphere which is  zero

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       p_1 =  [(0.5 * 1000 * (3.1)^2) +(0.008 * 1000 * 9.8) + 0]-

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