Question

Chapter 3, Problem 029 The figure is a section of a conducting rod of radius R1 = 1.40 mm and length L = 12.50 m inside a thin-walled coaxial conducting cylindrical shell of radius Ry - 12.4R, and the (same) length L. The net charge on the rod is +3.54 x 10-12 C; that on the shell is Q2 = 2.26Q. What are the (a) magnitude and (b) direction (radially inward or outward) of the electric field at radial distance r = 2.07R,? What are (c) E and (d) the direction at - 5.30R,? What is the charge on the (e) interior and (1) exterior surface of the shell? Units Units (a) Number (b) (c) Number (d) (e) Number (1) Number Units Units

Please be careful with negative positive and calculations

Answer 1

Given diagram is - -coloa-an - 3 Top view Liven Ri= trumm or 1-4X10m R2= 12.4 RI= 17-36 mon or 17.36X10m 11= 12-50m , 12312-sang Q1 = 3-5Uxiolac023-2426612-22-8.00uxió? at R r = 2.07R2 = 2:07x17 3 68153 of or r= 35.9352x10 sm we have to find electric field | 134 gauss #lawo $ Budisa cemen for r= 2.0782, a enco= alto 2-Q1 QI deneaz ar-cai- )= alene = as an so ( ECZKY lor (E)(22435-9352 x1 o 3x12.5) = 8.00uxio 23-surto? Eo (S] C28209132 416-3)= - Plasuuriosa S-854xio2 (09 (2820-9132X163) = -1-3038 LE= -6.UG 21 år! | 80 magnitude of E = TEL= 0nUG21 Vion

Direction of electric field is radially inwared. at r=5.3R1=5.3xl-uxio San - r = 7.42x15 Bom Cie in between cylinders by gauss daw - Bods = Qen. co Qene.= a1 = 3.54x1512 (E) (2AX742x163x12:5)= 3.suxoy so 8.8suxeste ( E 882.42X153) = 0.39981 so = 0.6864 ar so El=0.6864 vlam n is radially outward. By faradey's law equal but opposite natwee charge will incluce at interior surface of outer cylincher.i.e-a by change conservation law Q2-ch charge is on outer surface. so - Qiz -B-sux152c El ce-cei= - Biroux 151 – 3.sux10 12 Coa-con=-llasuuxio 12c