Question

(20 pts) A thick, infinitely long cylinder, with radius R is uniformly charged with volume charge density p. Using Gauss's Law, find the electric field for (a) r < R, and (b) r > R. P R

Answer 1

Solution Solid cylinder is Uniformly charged So Electric field haussiah. would be present surface inside as well as outside. The cylinder is long enough to Neglect the Edge Effect. Hence the È field due to cylinder is only Normal to the surface [ curve surface]- you can find direction first. charge density jo alma it can be converted into linear Charge den sidy do PX CSA surface area X = pxTRO NOW px circular For Finding Ê field inside at a distance r from center let us take a gaussian surface in form of cylinder of length e and radius r. You can see the diagram .I have the area. Now the High lieghted Gaussian Surface has three areas which are Named as 51, 52, 53, with theis vectors. area

E2l52 E3 NOW 53 an surface QY2 Q = tatal Here S, and Se are circular surfaces while S3 is CSA.[ curve surface] o charge Enclosed by Gaussian surface = O= volume charge E density X Volume of aussi SI Q xarze where AR²e charge NOW using crauss BE GO $£2.52 +683.53 GOR? as Ez=0 No parallel component only which is E3 15 present Hence also constant . otot Fз ф 53 [ as Ez and 33 have same direction p=00] R? Law En + =) & E.5 ER2 Q72 Ez x anre ERZ Ez= Qr2 2nGoreR2= 27 R²e&o pr Ang 260 n E

Now any point Enclosed For r7R total charge would be =Q EDxDRze -0 So Same type of Gaussian surface would be Chousen where r>R So using hauss Law Ę Dar²l E15, +6F2.5 +6 F3.5 Go Ez Ez - otot & 2nre DAR²e Go Рак?! neho PR² 3 26or E S2 27 PR Emax- 200 E x x inside F outside al E & 8