Please help me solve B and C and give an explanation. Thank you in advance. The
ID: 1533700 • Letter: P
Question
Please help me solve B and C and give an explanation. Thank you in advance.
The combustion of fossil fuels produces micron-sized particles of soot, one of the major components of air pollution. The terminal speeds of these particles are extremely small, so they remain suspended in air for very long periods of time Furthermore, very small particles almost always acquire small amounts of charge from cosmic rays and various atmospheric effects, so their motion is influenced not only by gravity but also by the earth's weak electric field. Consider a small spherical particle of radius r. density p. and charge q A small sphere moving with speed upsilon experiences a drag force F_drag = 6 pi eta gamma upsilon. where eta is the viscosity of the air. (This differs from the drag force you learned in Chapter 6 because there we considered macroscopic rather than microscopic objects.) A particle falling at its terminal speed upsilon_term is in dynamic equilibrium with no net force. Use Nev/ton's first law for this particle falling in the presence of a downward electric field of strength E to find an expression for upsilon_term. Express your answer in terms of the variables q. E. p. eta, r, and free fall acceleration g. Soot is primarily carbon, and carbon in the form of graphite has a density of 2200 kg/m^3. In the absence of an electric field, what is the terminal speed in mm/sofa 2.5 mu m-diameter graphite particle? The viscosity of air at 20 degree C is 1.8 times 10^-5 kg/m s. Express your answer using two significant figures. The earth's electric field is typically (150 N/C, downward) In this field, what is the terminal speed in mm/s of a 2.5 mu m-diameter graphite particle that has acquired 270 extra electrons? Express your answer using two significant figures.Explanation / Answer
r = d/2
= 2.5/2
= 1.25 micro m
B)
In the obsense of elctric field,
V_terminal = (4/3*pi*r^3*rho)/(6*pi*n*r)
= (4/3*pi*(1.25*10^-6)^3*2200)/(6*pi*1.8*10^-5*1.25*10^-6)
= 4.24*10^-5 m/s
= 0.042 mm/s
C) when elctric field exists,
V_terminal = (4/3*pi*r^3*rho + q*E)/(6*pi*n*r)
= (4/3*pi*(1.25*10^-6)^3*2200 + 270*1.6*10^-19*150)/(6*pi*1.8*10^-5*1.25*10^-6)
= 5.8*10^-5 m/s
= 0.058 mm/s