Reference. Vector Analysis Chapter 2


 Job Holland
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1 Reference Vector nlsis Chpter Sttic Electric Fields (3 Weeks) Chpter 3.3 Coulomb s Lw Chpter 3.4 Guss s Lw nd pplictions Chpter 3.5 Electric Potentil Chpter 3.6 Mteril Medi in Sttic Electric Field Chpter 3.8 Boundr conditions for Electrosttic Fields Chpter 3.9 Cpcitnce nd Cpcitors Intro.1
2 III. VECTOR NLYSIS 3.1 Vector lgebr ddition ssocitive lw +(B+C) (+B)+C commuttive lw +B B+ multipliction b sclr B B distributive lw ( B+C) B + C Intro.
3 Sclr (or Dot) Product:.B B..B B cos θ B.(B+C).B +.C Vector (or Cross) Product: B B sin( θ ) n B B n B θ B Intro.3
4 Intro.4 C B) C) (B B B B ( Note for cross products, Crtesin Coordinte Sstem (unit vectors ) B B B + +.B B B B B
5 Differentil displcement vector dl dl d + d + d dl Emple: To integrte F long the pth from (1,1) to b (,4), where F + b F.dl b ( + )( d + d ) + 1 d 4 d Intro.5
6 Differentil surfce vector ds: +ve in the outwrd norml direction to the surfce element for n enclosed volume Emple: Find F.dS for the surfce BCD for the given F in the following unit cube. ds B F C D Intro.6
7 Intro ) ( ) ( dd d dd d S F. S F d Note the minus sign in ds becuse positive direction is outwrd from the enclosed volume for surfce BCD
8 Intro.8 Clindricl coordinte sstem + 1 tn φ ρ ϕ ρ φ ρ ρ φ φ ρ
9 Intro r 1 1 tn cos φ θ Sphericl coordinte sstem θ r ϕ r r r φ θ θ φ φ θ
10 3. Sclr nd Vector Clculus Integrls of sclrs nd vectors over volumes, surfces nd lines often used in electromgnetics. Emples: ρ v is the chrge densit (per unit volume) D is the electric flu densit (or electric displcement) E is the electric field intensit V S C ρ v dv D.dS E.dl Totl chrge enclosed within volume V Totl electric flu pssing through the surfce S Potentil difference between two points on the line Intro.10
11 Grdient of sclr field f grd f f f + f + f grdf is vector in the direction of mimum increse of the field f. f f l m n n is unit vector in the direction of mimum increse of f Intro.11
12 Divergence of vector field : div lim v 0 S. d v S If we consider s flu densit (per unit surfce re), the closed surfce integrl represents the net flu leving the volume v In rectngulr coordintes, div. + + Intro.1
13 Divergence Theorem: If is vector, then for volume V surrounded b closed surfce S, V. dv S. d S The bove integrl represents the net fle leving the closed surfce S if is the flu densit V S Intro.13
14 Curl of vector field: Curl of vector is mesure of the tendenc of to push or pull round closed pth tht encircles point. Component of curl in direction i ( curl ) i ( ) i lim S i 0 C. i S d i l S i C i i Intro.14
15 Intro.15 In rectngulr coordintes m 0 lim S d C S l. n Since the mimum vlue of n component of vector is equl to the mgnitude of the vector,
16 Stokes s theorem: For n open surfce S bounded b contour C, S ( C ). d S C. d l S The line integrls from djcent cells cncel leving the onl the contribution long the contour C which bounds the surfce S. Intro.16
17 IV. ELECTROSTTIC FIELDS (TimeInvrint) The electric field intensit E is defined s the force on n unit positive chrge t point. D is the electric flu densit (or electric displcement densit). The direction of D is tht of the electric flu t the point, nd its mgnitude is the no.of flu lines pssing through n unit norml surfce re. 4.1 Guss s Lw :.D ρ Point form D. ds S v Q Integrl form Intro.17
18 ρ v is the spce volume chrge densit (coulomb per unit volume) t point. Q is the totl chrge enclosed within closed surfce S. Unit of D is C/m. Reltion between D nd E: D o r ε ε E εe ε o is permittivit of free spce (vcuum), ε r is the reltive permittivit (or dielectric constnt) of the medium mteril, nd ε is the permittivit of the medium mteril. (Unit of ε o nd ε is Frd/metre) In free spce, ε r 1 D ε E o Intro.18
19 Guss s lw cn determine the electric field pttern due to n distribution of chrges. In the simplest cse, consider the field t point P of distnce r from point chrge of mgnitude Q in free spce. +Q r P Construct sphericl surfce with centre t Q nd rdius r. E is the sme everwhere on the surfce. ppling integrl form of Guss s lw gives 4πr E Q ε o E Q 4πε r o r Intro.19
20 The force cting test chrge q t point P is then (Coulomb s lw): 1 4πε o qq r F r Emple : Find the electric field due to sphericl chrge distribution of rdius with uniform volume chrge densit ρ v in free spce. Gussin surfce of rdius r Intro.0
21 Solution: Cse (i): For Gussin surfce of rdius r less thn the rdius of the chrge distribution, the totl chrge enclosed b the Gussin surfce is: 4 Q π r 3 ρ v 3 ppling Guss s Lw to the Gussin surfce, Q 4π r E ( r ) ε E ( r ) rρ 3ε Cse (ii): For Gussin surfce of rdius r greter thn : Q 4 3 o r π 3 o ρ Intro.1
22 ppling Guss s lw, 4π r E ( r ) E ( r ) 3ε 3 o ρ r Q ε o r 4. Electric (electrosttic) potentil V: The electrosttic field is conservtive, i.e. E 0 E. dl 0 C ccording to vector identit, for n sclr V ( V ) 0 Intro.
23 So we cn define sclr electric potentil V which is esier to work with, (Hence unit of E is volt/metre) Integrting long pth E V b V V b E. dl Note tht the integrl is independent of the pth tken. Phsicll the potentil difference is equl to the work done in moving n unit chrge from point to point B. Potentil due to N point chrges: V 1 4πε o N Q k k 1 r r k Intro.3
24 4.3 Cpcitnce Cpcitnce between two conductors is defined s: C where Q is the chrge on the conductor nd V is the potentil difference between the conductors. Q V Intro.4
25 Emple: Find the potentil difference between two coil cclinders of rdius, b in the following digrm. ssume the surfce chrge densit (per unit re) on the inner conductor is ρ s. Hence find the cpcitnce per unit length of the clinder. Solution: Construct Gussin clindricl surfce of unit length nd rdius r. The Efield t rdius r is: ρ s r b πr E( r) ρ s E( r) ε r ρ s π ε r Intro.5
26 The potentil difference V b between outer nd inner conductor: V b V V b b E ( r). dr ρ sdr ρ s b ln ε r ε b Cpcitnce per unit length πρ s C Vb πε b ln Intro.6
27 4.5 Dielectric mteril Dielectrics re insulting mterils, nd contin bound chrges which cnnot move freel to generte currents. Bound chrges cn move short distnces under n electric field to form electric dipoles. ( dipole is pir of equl nd opposite chrges (+Q,Q) seprted t smll distnce r. The dipole moment p is equl to Qr r. ) p t E +ve ve Zero field Under field Intro.7
28 From mcroscopic view point, ll the minute dipole moments p k dd up to result in net dipole moment per unit volume clled the polrition vector P (C/m ): P lim v 0 N v k 1 p v k volume v where N is the number of dipoles per unit volume. The polrition (or bound) chrges cn lso contribute to the electric field E s the free chrges. (Note ρ v defined before is free chrge densit. ) It cn be shown tht D is relted to E nd P ccording to the following eqution: p k Intro.8
29 D ε E + o P In most mterils, P is in the sme direction nd proportionl to E (P ε o χe), so tht D ε E o + o r ε o χe ε ε E εe where ε o is the permittivit of vcuum, ε r is the reltive permittivit (or dielectric constnt) of the mteril, ε is the permittivit of the mteril, nd χ is the susceptibilit of the mteril. Intro.9
30 4.6 Presence of conductor Electric current will flow in conductor when there is n electric field. The current densit J (/m ) is proportionl to the electric field intensit E. J σe where σ is the conductivit of the mteril. This is nother form of Ohm s Lw. The totl current I flowing through surfce S is given b: I S J. ds In the interior of perfect conductor, σ is ver lrge so tht E cn be ssumed to be ero. Intro.30
31 END Intro.31
32 D θ ds D. d S D ds cos θ If D is the flu densit (per unit surfce re), D.dS represents the flu leving the surfce element ds in the norml direction. So S ds D. is the totl flu leving the entire surfce S. Intro.3
33 The phsicl mening of the integrl form of Guss Lw: the totl electric flu leving closed surfce S D.dS is equl to the net chrge enclosed b the surfce. The net chrge is just equl to the summtion of ll the positive or negtive chrges inside. Note tht the vrition of the positions of the chrges does NOT lter the totl flu leving the surfce; the vrition onl ffects the distribution pttern of the electric lines of force. Intro.33
34 n electrosttic field is conservtive, i.e. 0 E.dl C It mens the energ required to move n chrge round closed loop is ero! The reson is tht in moving long the pth, sometimes work is done b the electric force on the chrge, nd sometimes work is done ginst the electric force. If we regrd E.dl s the potentil difference between two points on the pth, E.dl must be ero becuse it is the potentil difference between the sme points. C Note tht E.dl is not necessril ero in timevring mgnetic field becuse of the induced e.m.f. due to the mgnetic flu. C Intro.34
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