EM Dis Ch 5 Part 2.Pdf

EM Dis Ch 5 Part 2.Pdf

1 LOGO Chapter 5 Electric Field in Material Space Part 2 iugaza2010.blogspot.com Polarization(P) in Dielectrics The application of E to the dielectric material causes the flux density to be grater than it would be in free space. D oE P P is proportional to the applied electric field E P e oE Where e is the electric susceptibility of the material - Measure of how susceptible (or sensitive) a given dielectric is to electric field. 3 D oE P oE e oE oE(1 e) oE( r) D o rE E o r r1 e o permitivity of free space permitivity of dielectric relative permitivity r 4 Dielectric constant or(relative permittivity) εr Is the ratio of the permittivity of the dielectric to that of free space. o r permitivity of dielectric relative permitivity r o permitivity of free space 5 Dielectric Strength Is the maximum electric field that a dielectric can withstand without breakdown. o r Material Dielectric Strength εr E(V/m) Water(sea) 80 7.5M Paper 7 12M Wood 2.5-8 25M Oil 2.1 12M Air 1 3M 6 A parallel plate capacitor with plate separation of 2mm has 1kV voltage applied to its plate. If the space between the plate is filled with polystyrene(εr=2.55) Find E,P V 1000 E 500 kV / m d 2103 2 P e oE o( r1)E o(1 2.55)(500k) 6.86 C / m 7 In a dielectric material Ex=5 V/m 1 2 and P (3a x a y 4a z )nc/ m 10 Find (a) electric susceptibility e (b) E (c) D 1 (3) 10 (a)P e oE e 2.517 o(5) P 1 (3a x a y 4a z ) (b)E 5a x 1.67a y 6.67a z e o 10 (2.517) o (c)D E o rE o rE o( e1)E 2 139.78a x 46.6a y 186.3a z pC/m 8 In a slab of dielectric material for which ε =2.4εo , V=300z2 V Find (a) D and ρv (b) P and ρpv (a) E V 600Z az 2 D E (2.4 o)(600Z)az 12.75 Za z nC/m 3 v D 12.75nC/m (b)P e oE (1 r) oE (1 2.4) o(600Z)az 2 7.43 Za z nC/m 3 pv P 7.43nC/m 9 A spherical dielectric shell with ε =εoεr for a<r<b and ε =εo for 0<r<a If a charge Q is placed at center of the shell find: Find (a) P for a<r<b (b) ρpv for a<r<b (a ) for a r b QQ DE22 4 rr4 o r DEP o QQ PDE oo 22 4 rr4 o r Q 1 P 2 1 4 r r 10 (bP ) pv 0 Two point charges when located in free space exert a force of 4.5 μN on each other. When the space between them is filled with a dielectric material, the force changes to 2 μN. Find the dielectric constant of the material Q1Q2 F1 4 r 2 o F2 Q1Q2 r 2 4 o rr F1 4.5106 2.25 r F2 2106 11 A conducting sphere of radius 10 cm is centered at the origin and embedded in a dielectric material with ε =2.5εo . If the sphere carries a surface charge of 4 nC/m2,find E at (-3cm,4cm,12cm). from Gauss's law : Q D.dS dS en s r 2 x2 y 2 z 2 r (3)2 (4)2 (12)2 13cm 2 2 D(4 (0.13) ) s (4 (0.1) ) 9 D 2.36710 E 2.5 oE 2.367109 E 106.92 V/m 12 2.5 o A sphereradius=a, εr ,uniform charge density of ρo. (a) At the center of the sphere, show that 2 oa V (2 r1) 6 o r V E.dl (1) for r a , o r (2) for r a , o Q D.dS dS Q D.dS dS en s en s 2 4 3 2 4 3 D(4 r ) ( r ) D(4 r ) o ( a ) o 3 3 3 or oa D a D ar 3 r 3r 2 ρ r ρ a3 E o a E o a r 2 r 13 3εo εr 3εo r a ρ r ρ a3 V E.dl o dr o dr 2 0 3εoε r a 3εo r ρ a2 ρ a3 1 o o [ ] 6εoεr 3εo a 2 ρoa 1 2εr 6εoεr (b) Find the potential at the surface of the sphere. ρ a3 V E.dl o dr 2 a 3εo r ρ a2 o 3ε r 14 LOGO iugaza2010.blogspot.com 15 [email protected].

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