(III) 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 R a and R b ) is used to create ionized air molecules for use in charging dust and soot particles (Fig. 24–21). Under standard atmospheric conditions, if air is subjected to an electric field magnitude that exceeds its dielectric strength E S = 2.7 × 10 6 N/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 R b = 0.10 mm and R a = 10.0 cm. In order to create a corona discharge region with radius R = 5.0 R b , 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.] FIGURE 24–21 Problem 20.
(III) 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 R a and R b ) is used to create ionized air molecules for use in charging dust and soot particles (Fig. 24–21). Under standard atmospheric conditions, if air is subjected to an electric field magnitude that exceeds its dielectric strength E S = 2.7 × 10 6 N/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 R b = 0.10 mm and R a = 10.0 cm. In order to create a corona discharge region with radius R = 5.0 R b , 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.] FIGURE 24–21 Problem 20.
(III) 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 (Fig. 24–21). Under standard atmospheric conditions, if air is subjected to an electric field magnitude that exceeds its dielectric strength ES = 2.7 × 106 N/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 = 0.10 mm and Ra = 10.0 cm. In order to create a corona discharge region with radius R = 5.0 Rb, 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.]
(i) Use Gauss’s law to find the electric field due to a uniformly charged infinite plane sheet. What is the direction of field for positive and negative charge densities?
(ii) Find the ratio of the potential differences that must be applied across the parallel and series combination of two capacitors Cj and C2 with their capacitances in the ratio 1 : 2 so that the energy stored in the two cases becomes the same.
(II) The electric field between the plates of a paper-separated
(K = 3.75) capacitor is 8.24 x 104 V/m. The plates are
1.95 mm apart, and the charge on each is 0.675 µC. Determine
the capacitance of this capacitor and the area of each plate.
A parallel-plate capacitor with plate area A = 2.0 m² and
plate separation d = 3.0 mm is connected to a 35-V battery
(Fig. 17–51a). (a) Determine the charge on the capacitor, the
electric field, the capacitance, and the energy stored in
the capacitor. (b) With the capacitor still connected to the
battery, a slab of plastic with dielectric strength K = 3.2
is placed between the plates of the capacitor, so that the gap is
completely filled with the dielectric (Fig. 17–51b). What are
the new values of charge,
electric field, capacitance,
and the energy stored in
the capacitor?
A = 2.0 m²
}d = 3.0 mm
35 V
(a)
FIGURE 17-51
35 V
K= 3.2 3.0 mm
Problem 96.
(b)
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