.. Electrostatic precipitators use electric forces to remove pollutant particles

Question

.. Electrostatic precipitators use electric forces to remove pollutant particles from smoke, in particular in the smokestacks of coal-burning power plants. One form of pre- cipitator consists of a vertical, hollow, metal cylinder with a thin wire, insulated from the cylinder, running along its axis (Fig. P23.67). A large potential difference is established between the wire and the outer cylinder, with the wire at lower potential. This sets up a strong radial electric field directed inward. The field produces a region of enters the precipitator at the bottom, ash and dust in it pick up electrons, and the charged pollutants are accelerated toward the outer cylinder wall by the electric field. Suppose the radius of the central wire is 90.0 mm, the radius of the cylinder is 14.0 cm, and a potential difference of 50.0 kV is established between the wire and the cylinder. Also assume that the wire and cylinder are both very long in comparison to the cylinder radius, so the results of Problem 23.63 apply.

(a) What is the magnitude of the electric field midway between the wire and the cylinder wall?

(b) What magnitude of charge must a 30.0-mg ash particle have if the electric field computed in part (a) is to exert a force ten times the weight of the particle?

Explanation / Answer

a) Gauss's law says that the total field through a closed surface is proportional to the charge contained by the surface. In this case, symmetry says that the E field should be radially directed, so the E field at a radius R in a cylinder L long is
Q / e0 = (2pi R E) L
E = (Q/L) / (2pi e0 R)

where Q is the total charge enclosed by the cylinder. The force on a particle with charge q is
F = qE
and the potential difference qV between one radius and another is the integral Fdx between the two radii:
qV = integral( q (Q/L) / (2pi e0 R) dR) = - [ q (Q/L)/(2pi e0) ] (1/R2^2 - 1/R1^2)
V = 50e3 V = - [ (Q / 14e-2 m)/(2pi e0) ] (1/(14e-2 m)^2 - 1/(90e-6 m)^2)

Solve for Q, then substitute to find E at R = (14e-2 m + 90e-6 m) / 2

b) The weight is mg = (30e-6 g)(9.8 m/s^2), so solve F = qE = 10mg for q

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