Question

9. A perfect conductor occupies the space y <0, as shown in the Fig. P9. There is no space charge or current in the air (Ho, )-filled space above the conductor (y > 0). (Points: 10) (1) Given the H-field in air just above a perfect conductor to be H, - & + 21 A/m, find the surface current), on the surface of the perfect conductor; (2) Given the surface charge density on the surface of the perfect conductor to be plynot = 1 nC/cm, find the E-field in air above the surface: E =? What is the direction of the corresponding E-field E, in air? Indicate the direction, amplitude, and unit for the electric field. Fig. P9

Answer 1

Solution :- Given the H-field tn aii just above a perfeel- Conductos to be. H. = +22 a lm, Griven fiqusle is •a<h,' Perfect/ , /comsluctool, Â = +22 4lm. Sur face =o. hence both part ot H, be, normal to it we know that ields n fole Condluctos are zero', AS 1-e., Eg -o, M2=0. an x (H)=Jc Csurface current), Ana

Whesle is Dormal Vecto9 lich s outward nomal an-a Hence ayx H, =3. -t21 Hence Surface Cerrent * = (2) Boundry Condi bin fog E field at perfect Cendluctoy DA = Ssyzt nclcima D, is eletric flun desity and it is whesle Normal to Scer face, hence Do Ps directed to -ay or) 0, Hence D,= E E, kle Koow tħat E is air's permibivity and E is Elketie fleld. Where -lX169 DI -11.299 Kvlcm Ei Eo 8-85 X164 E = -11:3 kylem.y
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