A small piece of dust of mass m = 2.5 µg travels through an electric aircleaner

Question

A small piece of dust of mass m = 2.5 µg travels through an electric aircleaner in which the electric field is 447 N/C. The electric force on the dust particleis equal to the weight of the particle. (a) What is the charge on the dustparticle?
1 C

(b) If this charge is provided by an excess of electrons, how manyelectrons does that correspond to?
2 electrons (a) What is the charge on the dustparticle?
1 C

(b) If this charge is provided by an excess of electrons, how manyelectrons does that correspond to?
2 electrons

Explanation / Answer

The weight of the particle isF=m*a=2.5*10-6*9.8=2.45*10-5 N Since the electric field had a strength of 447N/C, the charge musthave been: 2.45*10-5/447=5.48*10-8 C the charge of an electron is 1.6*10-19 C, so the totalnumber of electrons is5.48*10-8/(1.6*10-19)=3.43*1011electrons

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