20075ee1_1_HW6 - CHAPTER 6 6.1 Atomic hydrogen contains 5.5...

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CHAPTER 6. 6.1. Atomic hydrogen contains 5 . 5 × 10 25 atoms/m 3 at a certain temperature and pressure. When an electric field of 4 kV/m is applied, each dipole formed by the electron and positive nucleus has an effective length of 7 . 1 × 10 19 m. a) Find P: With all identical dipoles, we have P = Nqd = (5 . 5 × 10 25 )(1 . 602 × 10 19 )(7 . 1 × 10 19 ) = 6 . 26 × 10 12 C / m 2 = 6 . 26 pC / m 2 b) Find r : We use P = 0 χ e E , and so χ e = P 0 E = 6 . 26 × 10 12 (8 . 85 × 10 12 )(4 × 10 3 ) = 1 . 76 × 10 4 Then r = 1 + χ e = 1 . 000176 . 6.2. Find the dielectric constant of a material in which the electric flux density is four times the polarization. First we use D = 0 E + P = 0 E + (1 / 4) D . Therefore D = (4 / 3) 0 E , so we identify r = 4 / 3 . 6.3. A coaxial conductor has radii a = 0 . 8 mm and b = 3 mm and a polystyrene dielectric for which r = 2 . 56. If P = (2 ) a ρ nC / m 2 in the dielectric, find: a) D and E as functions of ρ : Use E = P 0 ( r 1) = (2 ) × 10 9 a ρ (8 . 85 × 10 12 )(1 . 56) = 144 . 9 ρ a ρ V / m Then D = 0 E + P = 2 × 10 9 a ρ ρ 1 1 . 56 + 1 = 3 . 28 × 10 9 a ρ ρ C / m 2 = 3 . 28 a ρ ρ nC / m 2 b) Find V ab and χ e : Use V ab = 0 . 8 3 144 . 9 ρ = 144 . 9ln 3 0 . 8 = 192V χ e = r 1 = 1 . 56 , as found in part a . c) If there are 4 × 10 19 molecules per cubic meter in the dielectric, find p ( ρ ): Use p = P N = (2 × 10 9 ) 4 × 10 19 a ρ = 5 . 0 × 10 29 ρ a ρ C · m 1
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6.4. Consider a composite material made up of two species, having number densities N 1 and N 2 molecules / m 3 respectively. The two materials are uniformly mixed, yielding a total number density of N = N 1 + N 2 . The presence of an electric field E , induces molecular dipole moments p 1 and p 2 within the individual species, whether mixed or not. Show that the dielectric constant of the composite material is given by r = f r 1 +(1 f ) r 2 , where f is the number fraction of species 1 dipoles in the composite, and where r 1 and r 2 are the dielectric
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