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# Note for ECE 2317 with Professor Jackson at UH dialectrics

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COURSE
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KARMA
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This 3 page Class Notes was uploaded by an elite notetaker on Friday February 6, 2015. The Class Notes belongs to a course at University of Houston taught by a professor in Fall. Since its upload, it has received 14 views.

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Date Created: 02/06/15
Electromagnetics Workshop Solutions Dielectrics Dielectric Breakdown Problem 1 Fused silica SiOz has a relative permittivity of approximately 38 The density of fused silica is 22 gcm3 There are 2144352X1024 molecules per kg of fused silica a If an electric eld of 10 Vm is applied to fused silica what is the dipole moment of each water molecule Assume that all of the molecules in the volume have the same dipole moment b Assuming that the charge q in the dipole formed by a single fused silica molecule is equal to four times the charge on a proton qw 1602 gtlt103919 C Calculate the effective displacement distance d between the positive and negative charges inside the fused silica molecule when the electric field inside the fused silica is 10 Vm Solution 2p P V 1280E8 16 0E Assuming all of the molecules in the volume have the same dipole moment we can simplify to N 1 EAV P sr 180E 3 pl g 1g0 EAV N We can arbitrarily choose AV 1 m3 Hence we need to find N the number of molecules in AV1m3 24 6 3 N 2144352 X 10 molecules 22 g 10 cm 1kg m 47176 X1027 moleculesm3 1kg cm 1 m 1000 g If p and E are aligned we can also write p1 28 lp W 5255x10 Cm dl MMm 8201gtlt1039 m q 461 Electromagneti cs Workshop Solutions Dielectrics Dielectric Breakdown Problem 2 Dielectric breakdown occurs in a material whenever the magnitude ofthe eldE exceeds the dielectric strength anywhere in the material a In a coaxial capacitor filledwith insulatingmatenal ofpeimittivity c at what value of p is E a maximum b What is the breakdown voltage if a 1 cm 17 2 cm and the dielectric material is mica with c o and Em zoo MVm Solution a From Gauss s Law we can easily show that 5 7L 27 e p Hence E is a maximum when p d t p p 17 p b VImdp7 lnilln2 EBR 2H 2 pl EBR 127rSua V l Wm 2 E aln 2 200x1051x10 2ln2sl386 MV 5quot IZIFSD 5quot Electromagnetics Workshop Solutions Dielectrics Dielectric Breakdown Problem 3 A metal sphere of radius a is in air The dielectric breakdown of the air is E Vm a What is the maximum total charge Q max that can be placed on the sphere before the air will break down b Assume that this charge is place on the sphere what is the potential on the surface of the sphere in terms of E5 Solution a By Gauss s Law E QL KZ p gt a 47230quot Breakdown occurs where eld is a max ie at surface r a E5 Q quot X 2 Vm 471396 061 So Qmax 47r50a2E5 C 2 b Note E 2E5 a 2Vm r gt a r r a JEdr jEdr j Qm dr QM 4 g a2Er 2 aEE Vm 471780 47178061 47178061

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