In an electrostatic air cleaner (\"precipitator\"), the strong nonuniform electr

Question

In an electrostatic air cleaner ("precipitator"), the strong nonuniform electric field in the central region of a cylindrical capacitor (with outer and inner cylindrical radii Ra and Rb) is used to create ionized air molecules for use in charging dust and soot particles (see the figure (Figure 1) ). Under standard atmospheric conditions, if air is subjected to an electric field magnitude that exceeds its dielectric strength ES=2.7×106N/C, air molecules will dissociate into positively charged ions and free electrons. In a precipitator, the region within which air is ionized (the corona discharge region) occupies a cylindrical volume of radius R that is typically five times that of the inner cylinder. Assume a particular precipitator is constructed with Rb = 0.20 mm and Ra = 20.0 cm . In order to create a corona discharge region with radius R=5.0Rb, what potential difference V should be applied between the precipitator's inner and outer conducting cylinders? [Besides dissociating air, the charged inner cylinder repels the resulting positive ions from the corona discharge region, where they are put to use in charging dust particles, which are then "collected" on the negatively charged outer cylinder.]

Explanation / Answer

The inner radius is:
Rb = 2e-4 (m)
The outer radius is:
Ra = 2e-1 (m)
The radius of the ionized region is 5 x Rb:
R = 5e-4 (m)

So if the inner cylinder has charge Q applied, the outer cylinder must have charge -Q. This imposes a voltage difference that can be found from Gauss' law:
Q/0 = E.da
= E(r) * 2r * L
E(r) = Q/(2rL*0) = (Q/(2L*0))/r

The voltage between r = Ra and r = Rb is therefore:
V = - E(r) dr = - (Q/(2L*0)) * dr/r = - (Q/(2L*0))*ln(Ra/Rb)

The sign is not important, so we will just set:
V = (Q/(2L*0))*ln(Ra/Rb)
Q = (2L*0)V/ln(Ra/Rb)

The electric field at the critical radius R is:
E(R) = (Q/(2L*0))/R
= (V/ln(Ra/Rb))/R

and according to the concept, this must be Es = 2.7 * 10^6 (N/C):
Es = E(R) = (V/ln(Ra/Rb))/R
V = R*Es*ln(Ra/Rb)

Therefore, the numerical value is:
V = 5e-4 * 2.7e6 * ln(2*10^-1/2*10^-4)
= 5*2.7*ln(1000) * e2
= 5*270*3*ln(10) = 9325.5 (V)

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