Question

5.19

od of radius 1.5 cm in sistivities of copper and 5.11 A composite conductor 10 m long consists of an inner core of steel of radi an outer sheath of copper whose thickness is 0.5 cm. Take the resistivities of steel as 1.77 X 10 and 11.8 X 10" . m, respectively. (a) Determine the resistance of the conductor (b) If the total current in the conductor is 60 A, what current flows in each metal? (c) Find the resistance of a solid copper conductor of the same length and cross-sectional areas as the sheath 5.12 The cross section of a conductor made with two materials with resistivities p, and p, is shown in Figure 5.19. Find the resistance of length of the conductor. A 12 V voltage is applied across the ends of a silver wire of length 12.4 m and radius 0.84 mm. Determine the current through the wire. 5.13 Sections 5.5-5.7--Polarization and Dielectric Constant 5.14 5.15 At a particular temperature and pressure, a helium gas contains 5 X 10 atoms/m. Ita 10 kV/m field applied to the gas causes an average electron cloud shift of 10 m, find the dielectric constant of helium. Adielectric material contains 2 x 10 polar molecules/m', each of dipole moment 1.8 X 10°C.m. Assuming that all the dipoles are aligned in the direction of the elec tric field E - 10'a V/m, find P and er A 10 mC point charge is embedded in wood, which has e - 4.0. Assuming that the charge is located at the origin, find Patr = 1 m. In a certain dielectric for which , = 3.5, given that P a nC/m'. find E and D at P = 2 m. 5.16 5.17 D_100 5.18 5.19 A cylindrical slab has a polarization given by P p pa... Find the polarization change density inside the slab and its surface charge density Pr A spherical shell has = 1.2 cm and = 26 cm as inner and outer radil, respective P = 4ra, pc/m, determine (a) the total bound surface charge on the inner surfaces the total bound surface charge on the outer surface, (c) the total bound vom

Answer 1

Given: The radius of the inner surface r = 1.2 am The radius of the outer surface r. = 2.6 cm = 4rx10-12a, Solution: The total surface bound density is given by, P= Pa = 4rx10aa P = 4rx10-2 ...... (1) Q=Pxx surface area of sphere = 47x10-2 x 472 = 167 X10-2 PART A The total bound surface charge on the inner surface is given by, Qotre =16x7x10-2 ...... (2) On substituting = 1.2 am in (2) Disne=16x x10-2 (1.2 x104) = 8.6858x10-C PART B The total bound surface charge on the outer surface is given by, Que=16xr; x10-2 ...... (3) On substituting , = 2.6 cm in (3) Oste=164 x10-12 x(2.6x10-) = 8.8346 x10-4C PARTC The total bound volume is given by, Q = Osse - One On substituting Que=8.8346 x10-4C and = 8.6858 x10" Cin(4) Quier = 8.8346x10-45-8.6858 x10-17 = 7.966x10-16C