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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

 r1 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 2103 2 P  e oE  o( r1)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( e1)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

DEP o QQ PDE oo 22  4 rr 4  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 4.5106     2.25 r F2 2106 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  2.36710  E  2.5 oE 2.367109 E  106.92 V/m 12 2.5 o A sphereradius=a, εr ,uniform charge density of ρo. (a) At the center of the sphere, show that 2  oa V  (2 r1) 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

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