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 precipitator consists of a vertical, hollow, metal cylinder with a thin wire, insulated from the cylinder, running along its axis. A large potential difference is established between the wire and the outer cylinder, with the wire at the lower potential. This sets up a strong radial electric field directed inward. The field produces a region of ionized air near the wire. Smoke 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 µm, the radius of the cylinder is 12.5 cm, and a potential difference of 59.00 kV is established between the wire and the cylinder.

(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 µg ash particle have if the electric field computed in part (a) is to exert a force ten times the weight of the particle?

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Answer #1

Electric field is, FVab 1 Eln(b/a) r 59000v In(0.125/0.00009) o.0625 E 130454.14 V/m E= 1.3 x 105 V/m 10*m*g -130454.14*q q=

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