LECTURE NOTES 2 - University Of Illinois

UIUC Physics 435 EM Fields & Sources IFall Semester, 2007Lecture Notes 2Prof. Steven ErredeLECTURE NOTES 2Gauss’ Law / Divergence TheoremConsider an imaginary / fictitious surface enclosing / surrounding e.g. a point charge (or a smallcharged conducting object). For simplicity, use an imaginary sphere of radius R centered on chargeQ at origin:ẑnˆ , r RrˆE ( r ) E ( Rrˆ )Infinitesimal Area Element, dAQ θ RŷϕImaginary/Fictitious Surface, Saka Gaussian Surface of radius Rcentered on charge Q.x̂SArea element dA is a VECTOR quantity: dA dAnˆ dArˆ . By convention, n̂ is outward-pointingunit normal vector at area element dA. In this particular case (because of spherical symmetry ofproblem): nˆ rˆFLUX OF ELECTRIC FIELD LINES (through surface S): Φ E E ( r )idASΦ E “measure” of “number of E-field “lines” passing through surface S, (SI Units: Volt-meters).) associated with any closed surface S, is a measure of the (total)TOTAL ELECTRIC FLUX ( ΦTOTEcharge enclosed by surface S.n.b. charge outside of surface S will contribute nothing to total electric flux Φ E (since E-field linespass through one portion of the surface S and out another – no net flux!)Consider our point charge Q at origin. Calculate the flux of E passing through a sphere of radius r:(see above picture)Q 1 2Φ E E ( r )idA rrˆ i r sin θ dθ dϕ rˆ S4πε o S r 2 () dAinfinitesimal vectorarea element forsphere of radius rn.b. Vector area element of sphere of radius, r is dA dArˆ ( r 2 sin θ dθ dϕ ) rˆ in spherical-polarcoordinates. Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois2005 - 2008. All rights reserved.1

UIUC Physics 435 EM Fields & Sources IThus: Φ E Q4πε oθ πϕ 2π o o θ ϕFall Semester, 2007sin θ dθ dϕ ( rˆirˆ ) 12π Q4π ε oθ π θ oLecture Notes 2Prof. Steven Erredesin θ dθ2 2Q Q 2εo εoΦE Gauss’ Law (in Integral Form):Q E ( r )idA εenclosedsoElectric flux through closed surface S (electric charge enclosed by surface S)/ ε oIf ( there exists) lots of discrete charges qi (ALL enclosed by imaginary / fictitious / Gaussiansurface S), we know from principle of superposition that:NENET ( r ) Ei ( r )i 1Then: Φ ENET NSENET ( r )idA i 1( S)Ei ( r )idA i 1qiεo 1εo qi 1i QenclεoIf volume charge density ρ (r ′) , then: Qencl ρ (r ′)dτ ′vThen using the DIVERGENCE THEOREM:ΦE Q E ( r )idA ( i E ( r ) ) dτ ′ εenclSvo 1εo ρ ( r ) dτ ′vThis relation holds for any volume v the integrands of Gauss’ Law (in Differential Form):2 i E ( r ) ρ (r ) ( ) dτ ′ must be equal, i.e.:vεo Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois2005 - 2008. All rights reserved.

UIUC Physics 435 EM Fields & Sources IFall Semester, 2007Lecture Notes 2Prof. Steven ErredeThe DIVERGENCE OF E ( r ) : i E ( r )Calculate i E ( r ) directly from E ( r ) 14πε orˆρ ( r ′ ) dτ ′r2 vallspacen.b. now extended over all space!Remember that r is NOT a constant!r r r′field sourcepoint pointPS 1 i E ( r ) i 4πε o 1 rˆ ′′ρrdτ v r 2 ( ) 4πε oall space rˆ Now: i 2 4π δ 3 ( r ) r 3 D vallspace rˆ i 2 ρ ( r ′ ) dτ ′ r (see equation 1.100, Griffiths p. 50)Diracδ fcn. rˆ Thus: i 2 4πδ 3 ( r r i E ( r ) 14π ε o ) r r′ or : i 4πδ 3 ( r r ′ ) r r′ 3 4π δ 3 ( r r ′ ) ρ ( r ′ ) dτ ′ vallspaceρ (r )εoGauss’ Law in Differential Form: i E ( r ) ρ (r )εoGauss’ Law in Integral Form: i E ( r ) ρ (r ), thus:εo ρ ( r′) 1 dτ ′ vεo εo ( i E ( r ′) ) dτ ′ V1 ρ ( r ′) dτ ′ εvQencloNow apply/use the Divergence Theorem on the volume integral associated with i E ( r ′ ) :11 ( i E ( r ′) ) dτ ′ E ( r )idA ε ρ ( r ′) dτ ′ ε QvSThus we obtain:voQ E ( r ′)idA′ εenclSencloGauss’ Law in Integral Formo Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois2005 - 2008. All rights reserved.3

UIUC Physics 435 EM Fields & Sources IFall Semester, 2007APPLICATIONS OF GAUSS’ LAWLecture Notes 2Prof. Steven Errede- very explicit, detailed derivation -Griffiths Example 2.2: Find / determine the electric field intensity E ( r ) outside a uniformly chargedsolid sphere of radius R and total charge q:draw concentric Gaussiansurface with radius r Rcentered on solid chargedsphere of radius R.ẑrˆ, nˆRrΟField Point P @ ron Gaussian surfaceŷInfinitesimal area elementdA dAnˆ dArˆdA r 2 d ( cos θ ) d ϕCharged solidSphere ofRadius R,Total charge qGauss’ Law: r 2 sin θ dθ dϕx̂Fictitious / Imaginary sphericalGaussian surface S of radius r1 E ( r )idA ε QenclSoE ( r ) E ( r ) rˆ 1εoq qεon.b. by symmetry of sphere:E sphere ( r R ) E ( r ) rˆdA dAnˆ dArˆ(for Gaussian sphere)i.e. E must be radial!! E ( r )idA ( E ( r ) rˆ )i( dArˆ ) E ( r ) dA ( rˆirˆ ) E ( r ) dA 1n.b. Here again, by symmetry,NOTE: E ( r ) E ( r ) the magnitude of E is constant (for all)/for any fixed r!!!(on the Gaussian spherical surface). q E ( r )idA E ( r ) dA εSSo E ( r ) dA E ( r ) ( 4π r 2 ) qSεoq1 qqrˆ rˆor: E ( r ) 224πε o r4πε o r4πε o r 24πε o r Electric field outside a charged sphere of radius R at radial distance r R from center of sphere. E (r ) q2 1n.b. the electric field (for r R) for charged sphere is equivalent / identical to that of a point charge qlocated at the origin!!! Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois42005 - 2008. All rights reserved.

UIUC Physics 435 EM Fields & Sources IFall Semester, 2007Lecture Notes 2Prof. Steven ErredeGAUSS’ LAW AND SYMMETRYUse of (Geometrical / Reflection) symmetry (and any / all kinds of symmetry arguments in general)can be extremely powerful in terms of simplifying seemingly complicated problems!! Learn skill of recognizing symmetries and applying symmetry arguments to solve problems!Examples of use of Geometrical Symmetries and Gauss’ Lawa) Charged sphere – use concentric Gaussian sphere and spherical coordinatesb) Charged cylinder – use coaxial Gaussian cylinder and cylindrical coordinatesc) Charged box / Charged plane – use appropriately co-located Gaussian “pillbox” (rectangularbox) and rectangular coordinatesd) Charged ellipse – use concentric Gaussian ellipse and elliptical coordinatese) Charged planar equilateral triangleThink aboutf) Charged pyramidthese!!APPLICATIONS OF GAUSS’ LAW (CONTINUED)- very explicit detailed derivationGriffiths Example 2.3 Consider a long cylinder (e.g. plastic rod) of length L and radius S that carriesa volume charge density ρ that is proportional to the distance from the axis s of the cylinder / rod –i.e. coulombs ρ(s) ks ( meter )3 coulombs k proportionality constant ( meter )4 a) Determine the electric field E ( r ) inside this long cylinder / charged plastic rod- Use a coaxial Gaussian cylinder of length l and radius s: (with l L)Gauss’ LawQ E ( r )idA εenclSEnclosed charge:oQencl ρ ( s′ )dτ ′ ( ks′ )( s′ds′dϕ dz ) integral over Gaussian surfacevQencl vs′ ss′ 0ϕ 2πz ls′ s 0z 0s ′ 0 ϕ ( ks′)( s′ds′dϕ dz ) 2π kl s′2 ds′2Qencl π kls 33 Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois2005 - 2008. All rights reserved.5

UIUC Physics 435 EM Fields & Sources ILHS Gaussian endcapSFall Semester, 2007Lecture Notes 2Prof. Steven ErredeCoaxial Gaussian cylinderlsnˆ zˆLong charged cylinder ofradius S and length Lnˆ zˆSẑz 0z lnˆcyl . rˆL(L l)RHS Gaussian endcapCylindrical Symmetry E ( r ) E ( r ) rˆ (i.e. E points radially outward, to z-axis.) E ( r )idA E ( r )idAcyl . Scylindricalportion ofGaussiansurface E ( r )idALHS endcapLHS endcapportion ofGaussiansurface E ( r )idARHSRHS endcapportion ofGaussiansurfaceendcapAgain, from cylindrical symmetry (here):E ( r ) E ( r ) constant on cylindrical Gaussian surface – i.e. fixed r r sWhat are dAcyl . , dALHS , and dARHS ?endcapdAcyl . sdldϕ rˆ ( nˆcyl . rˆ )endcapdALHS sdsdϕ ( zˆ ) sdsdϕ zˆ ( nˆ LHS zˆ )endcapinfinitesimal surface areaendcapdARHS sdsdϕ ( zˆ ) sdsdϕ zˆ ( nˆ RHS zˆ )element of Gaussian cylinderendcapendcap E ( r )idA ( E ( r ) rˆ )i( sdldϕ rˆ ) ( E ( r ) rˆ )i( sdsdϕ zˆ ) ( E ( r ) rˆ )i( sdsdϕ zˆ )Cyl RHSGaussianendcaprˆ i rˆ 1rˆ i zˆ 0ˆ 0rˆ i z Note(s):E ( r ) E ( r ) constant on cylindrical Gaussian surface (fixed r s)E ( r ) E ( r ) rˆ by symmetry of charged cylinderOn LHS and RHS endcaps E ( r ) is not constant, because r is changing there - (but E still pointsin r̂ direction! However, note that rˆirˆ 1 and rˆi( zˆ ) 0 Gaussian endcap terms do notcontribute!!!Constant here E ( r )idA SGaussiancylinder cylindricalGaussiansurfaceE ( r ) sdldϕ E ( r ) s Putting this all together now:6ϕ 2πz l ϕz 0Q E ( r )idA εenclSo 0dldϕ E ( r ) sl ( 2π ) 2π slE ( r )2where (here): Qencl π kls 33 Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois2005 - 2008. All rights reserved.

UIUC Physics 435 EM Fields & Sources IFall Semester, 2007Lecture Notes 2Prof. Steven Erredeinside32π s l E ( r ) 2π ks 2 l3ε oEin ( r ) or:(ks 2rˆ3ε o)n.b. r s as used in Griffith’s(s r S )book, page 73b) Find ELECTRIC FIELD E ( r ) outside of this long cylinder / charged plastic rodAgain, use Coaxial Gaussian cylinder of length l ( L) and radius s ( S):Q E ( r )idA εenclGauss’ Law:So2Qencl π klS 33Enclosed charge (for s S):coaxial Gaussian cylinderradius s S and length l Lnˆcyl . rˆLong charge cylinder ofradius S and length Lsnˆ LHS zˆnˆ RHSEndcap zˆSẑEndcapl LLAgain, from symmetry of long cylinder E ( r ) E ( r ) rˆ constant (radial) direction!!r s (fixed radius) E ( r )idA SE ( r )idAcyl cylindricalGaussiansurface E ( r )idALHS endcapLHSGaussianendcapE ( r )i dARHSendcapdALHS sdsdϕ ( zˆ ) sdsdϕ zˆ dALHSdAcyl sdldϕ rˆendcapendcapand( zˆ )endcapdARHS sdsdϕ ( zˆ ) sdsdϕ zˆ dARHS dAcyl rˆ dAcyl rˆNow: rˆirˆ 1 RHSGaussianendcap( zˆ )endcaprˆi( zˆ ) 0 Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois2005 - 2008. All rights reserved.7

UIUC Physics 435 EM Fields & Sources IFall Semester, 2007Then:Lecture Notes 2Prof. Steven Errede 0 E ( r )idA ( E ( r ) rˆ )i( dA rˆ ) ( E ( r ) rˆ )i anendcap E (r ) z lz 0ϕ 2π ϕ 0 0 zˆ RHS( E ( r ) rˆ )i dA RHSendcap zˆ endcapsdldϕ 2π slE ( r ) Electric field outside charged rod (s r S) : Eout ( r ) ELECTRIC FIELD (INSIDE/OUTSIDE)vs. radial distance sInside (s S):ks 2Ein ( r ) sˆ3ε o2π k l S 3kS 3r 3sε o3i 2π s l ε oLONG CHARGED CYLINDER(radius S, ρ(s) ks)Outside (s S):kS 3 1 Eout ( r ) sˆ3ε o s ( sˆ rˆ )Make a plot of E ( r ) vs. radial distance s:E (r )Emax ( s S )kS 2 3ε oVaries as s2Varies as 1/sRadialDistances08s S Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois2005 - 2008. All rights reserved.

UIUC Physics 435 EM Fields & Sources IFall Semester, 2007Lecture Notes 2Prof. Steven ErredeAPPLICATIONS OF GAUSS’ LAW - very explicit / detailed derivation –Griffiths Example 2.4:An infinite plane carries uniform charge σ (coulombs / meter2).Find the electric field a distance z zo above (or below) the plane.Use Gaussian Pillboxcentered on ge-on Perspective:ẑz h/2x̂ (out of page)hŷz h/2ly l/2y l/2Again, from the symmetry associated with -plane,E ( r ) E ( r ) zˆ E ( z ) zˆ (above plane), E ( z ) zˆ (below plane)The Gaussian Pillbox has 6 sides – and thus has six outward unit normal vectors: :nˆ5 , zˆA2 (back)nˆ2 , xˆA5 (top)nˆ4 , yˆnˆ3 , yˆA4 (RH side)A1 (front)A3 (LH side)nˆ1 , xˆnˆ6 , zˆA6 (bottom) Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois2005 - 2008. All rights reserved.9