Question

Problem 1) a) [10 marks] Volume charge is distributed in the cylindrical coordinate system the following charge densities: (r,9,z) with Py=+por Pr=0 Pr= Po forr <R , for R,<r<R2 , and for r>R2 00 where po>0, Ry, and R, are constants. Use Gauss' law to find Ē in all regions. Clearly indicate the choice of a proper Gaussian surface for each region and how the law is applied.

Answer 1

Problem 1 - E=permetiarity t3 CSR2 volume charge (Bu) for rLRI fu=tfor Sv=0 Sv=-So R and Rz for for Ricrera 8>R2 are constant. Gauss Law: It state that total flex crossing through Gausian surface es equal to total charge include by the aussian surface. Spods = o incloses - SSS Sv Differential - Volume Differential Total charge Area volume charge density inclosed in Gaussian Surface Electric flex Density D=EE - Electric field " Permitivity of mediem

calculating & ting Electric field in the region of real Cylinder having Radius Ri 9 1 z Height of cylinder cylindrical gausioan surface gaubian surface having radius using Gauss law in SS Dods = US svody DE EIE I Sv= for du= 8 do dodz I In cylindrical I coordinate Sds= 2182 Area of cylinder Casus face) & 22 z erordo do dz Do2arz = A 4E2932 = $o $*** So *ds. dz = Soppa To 10).* 123. = 3 (-) (27-09) (2-9 GE2h a so z izhx 24 z 8€ Bota

• E= 372 for Electric Electric & field when rer 3 EU • Calculating ® calceclating & e In the Region of R40 2 R2 outer cylinder Radius R2 having > cylindrical Gaussian surface radius r Inner cyliner Radiub R, Using causs Law: soods - SS Buch D= Ez E . I sua for JJds=2252 | Su=o NER, q2r< R₂ R za z > Daarz = sorordradda since brzo learer SO No charge inclosed between = 36 for dry add $dz = $. 1931 1932 12] E2 E 2/287 = % R3 x 24x7 a som im alm

= / Electric field for a co SoRi3 3tar 4 calculating eculating Electric fiels for 8> R2 Electric to Radius Re sylindrical Gaunian surface having radius cylinder Applying causs law in SS sroder Sloods= D= 63 € 1 calculated above 8v = 180 du= r do do dz =aarz 3 2 27 g - so fardd dat so Rozaz Daarz ↑ Total charge in cylinder R, it z =-30 / ar fad ļ dz + so Rs 272 eos = -8 [OJE 147* 12]. +372 242 = -f [P-R2) 2x 2 + 80 R13 282 Ez € 2tr7 = 25[ - 865-R3) + So RAST

- 3. ko-Rz) + 2) + BoRi Electric fiele Electric field when r> R2 - EBE Hence, Electric field (E) = for rari SR3 for p <r < R2 3 tar for ruR = -f(r-R2 Eze