Electromagnetics So Far as Oct 4 PDF

Sep 4 , 2024 Phys 5020 : Electromagnetics Eric Hessels Kessels @yorkv. Ca PSE 201 Problem Sets 10 % (posted and returned eclass) on 2 In-cla...

Sep 4 , 2024 Phys 5020 : Electromagnetics Eric Hessels Kessels @yorkv. Ca PSE 201 Problem Sets 10 % (posted and returned eclass) on 2 In-class tests 11 29 hour deadline : , rightdefore class. ↳ Test 1 25 % (October 7) ↳ Test 2 25 % Final exam 10 % Take look to for this weekend. * a Chapter 1 review And that's it for today. ( ?)  2024 Sep 6 , Coulombo's Law q'causes the force ↑ Q F - Q experiences the force Fi > q Force on Q : F = K q'Qi SI units : k = 43 12 · 50 = 8. 85x10-12 /Nm2 Superposition Add up charges of the system. & at F ; 9 92 at ,, , etc... =- > - F=Q If there is a line charge Charge per it length E Lime charge X de & charge ri Ex. [ g 2o 7 F r X y = = ' (x) y , 0) w Cuz , ay y x = - dq' X dl'Ade" =. Xo z)de 3Y & at (x , , y = z Figuring out del dX1 = du dl2 dxx + dyz = & = acudu de duw + 4022 =  the Setting up the components of integral F =t With ! 3 integrals : Ex , Fy , Fz  9 2024 Sep , Now for a surface S (Surface Charge) cr · (onQ) ps = a I > P Ex Sphere of radius R. V = r. COSO a E O O Q at F = (X , y, z) E R Set up · G Pick origin of sphere as O. g O = (X' y' , , z) = (Rsincos' , RsinG sing' , RCOSE) E0 R= dA Rsindd · = 8 O o : 0 COSO L. First 3 double integrals for each component. For a 3Dobject # (onQ) = ( · V We will be. setting up triple integrals for each component S Electric Field In all cases #x Q , write F = QE(F) on ↑ Q Electric field Fon E(r) Q Electric field : = Q E(r) = volume For a charge similar for Elr) = gridV , surface , line charge In general ECF) is/ : = charges dq do y DIRAC-DELTA n FUNCTION e What is g(r) for a point charge qf ?.S(r) (f) dV = g is zero away from point charge If a point charge had radius R.  11 2024 Sep , If we have a point charge of radius R. pers [1 0 (1 > R Using Dirac-Delta function Point charge p(r) zo. up(r) q5(x))5(y)5(z) = (v) = 0 except at = 0 Sig(r) dV = S J JodEq(xECyEL , Sphere = q(8(z -- de dy I 11 ro (40 zo) Point charge at yo = of , , g (r)) = q5(x) x)J(y y0)f(z - = - zo) = 5( ) q Surface charge Or On Z = zo plane (r) =(z' ) 50 & = - z 0 line Line chargeNo at X1 Xo, = Y' = Yo g(m) = X. 5(X) x)5(y = yo) - Units of F(x) /(dx = 1 > - units of [(x) : gth SE)dv = () :  Integrals becoming * easy ((x1)[(x) - x)dx) = f(x) g()()dV = g back to Electric Field Moving E(r) = 43 (r) dVi E is a vector field , a vector at each (vector) position I in space. E ? EX. We have a infinitely long straight line charge Xo. What is A - dz VS + dz' dEz E(r) )dEs d = gl)dv - = - - T Setting up dE des = 2 cost us zing2 De S 2x dz din = dq' = , COSO = (g2 + z2)"2 Xo Setting up the integral 2/ constant the defitzydex Tri S E = Ess a = E= cost SISYS)SEdO 53 Sec3O Z = Stand - dz' = SSecIO eE  15 , 2024 Sep (ii) 5 3"(i)h.. Recalling... E() = soSgd =- E field The is a divergence. V. E(r) = 45 JdV'g() o ↑ aPace set to Recalling the divergence 1 ↑ () ((X = - X((2 + (y y( - + (z - z1)2y32(X - x)) , (y y() (z - , - z) (X-X' (y y) = 27y((X - + [(X - X((2 + (y y( + (z z1)2j312 - - - X (2 + (y y( + (z / - - z1)27312 (z z) + - [(X - X((2 + (y y( + (z z1)27312 - - (y-y (z-z(2]-3(X-X) - 3(y-y-b(z-z). 8 32(x-x + [(X + X()z + (y y') (z z1727312 + O = For - - - 1 E V 0 X ~at F - &. = = it 1 Look at SV "DIVERGENCE THEOREM" St. dV = Godhards St = sphere sind od centered at in (R = 0) : G =Tododd = 4π ↑ = O for - E function St. d == 4) Plug in 4 SdVg(r) 48F =T We obtain the differential form of Gauss' Law - V · E() = g(r) Another Pathway St. E(F)dV = CdU FOR ANY"V Theorem - Qenclosed inside V Edi GE d Den. = Eo Any s int. form Gauss' Law Recalling...E. dE is the flux of Ethru S. Sin F x + -  Sep 16 , 2024 Recalling Gauss Law Ex. Infinite Line Charge. Finite GE d Gene G Surf : Cylinder ↳. : = Eo Eat top 5 & down Ed at bottom JEdA = O Seda & constant onS = Erder : Err j (5 ↓ = , 0, z) Ex. Finite Line Charge -↑ R Xo 5 GE. d = Seve E Always true , not always e we proceed with the inteded way. E(r) = de dq' Xjdz = Curl of E E= JdV'gire EXE = JdV'gir) x Xsin with respect to ir ij - xEx ByA [3312[7312 [3317 z - z -z()()(y-y-ily-y(2) = 2... 752 +Oj + Ok SoO. True even at = Stoke's Theorem de Hand Rule by Right ((x) : did out board ECr) Applying to =Ed - I O for any S True for any closed path c.  li ij sij ; is in his is Define electrostatic potential () [often (r)] - Ede Terrible definition ! D: W(r) = ~ Hessels , E. ↑ some reference point Along which path ? F(r) viCr) Var N 3 - di = -E. = ONYPATH · Sep 18 2024 , Recalling V(F) = de (NOTE Often : Scalar field * V (r) is not potential energy , but related. Potential Differences (B/wa 1) , ↑2 points ~ (6) - V(a) = -G Any patio Fundamental Theorem of Gradients * ~ (b) - ~(a) = Strode For a o Ex. Jinxdx Ssxdx = > - sinX = COSX f(xdx xx f(x) For all = = g(x) > - Integrands equal E = -V-V-  EIf Expontchageqatsompoint 1 EIF) = at If =0: Elr) = Vir = -E. Any path de · · Pick O at path : straight X line ~ di setting d = - up /d) de dr = dr ↑ S Negative de = dr - within the plug in integral ~ () =_ G -d = EX. o at general point I = -(r) = * Check to get -V Ex. Infinite line charge X. We had : E = 525 Cr) = -S ↑ ds - at So from X VIr) = - is d 2s = o = [s]ns-is zaxis Electrostatic Potential and Energy Refx. Compute energy required to move charged particle Q from a to Energy= work. Recall W = S. di F QE = should be negative to counteract EL force..: - DE 1. d = - /Edi = QU(1) - QU(b) Ind. path So QV(r) is a potential energy for Q The fact that this is independent of path , we have a conservative force ,  Sep 20 , 2024 Recalling... Work W = Q V (F) Work to and (N) = 0 Q (E) acts like a potential energy so Ur) < like ↑ Urgh Pot. energy EX. How much work is required to move Q from one point 2m away from an infinite line charge X to a point Im away. n Work = Q [V (b) - v(a)] - 3 In ( = = Using superposition For V(F) , n charges E(r) = -"El VV(F) = V(r) is distributions g(r) For cont charge :. V(r) due to g(r) dV' + add up = - -M 1) Bra = a = ~ (r) , E(r) = - VV(r) EX. We have an spherical surface uniformly charged with a total charge Q We. would like to know what V (F) is. Set up g(r) & R & It) = 2 surface but , we need to express in : gir) = E(r' R) - L Set up the integral for [T(F) W (8) = 3 SI(r' - R) dV' set integral (triple) and limits of We have , so we up a volume , integration. 44OsinGdOdq'dr'E Ur-ERCOSO Careful with the expression Don't forget theime.. g() = sineSind' do UR2 + r2-2RrCOSE" U du = = R2 + r2 erRCOSE' - ZuRsinde or ver du = 1/2 - [ For USR VIrl =REIR] Cr = Like point Q at 0. ForaR W(r) = ungn[r] =p const. E(8) = ww(r) - ~R : For r < R E = c  2024 Sep 23 , Conductors Perfect conductors have zero resistance. The electronsare free to move. We would say that - Perfect conductor : Perfectly free· ( Inside the conductor DE = 0 : if not feels force + more redistribute until E 0 =. E 0 ② = 0 : v. = ③v = constant : - V= E ① If the conductor is charged charge , is on the surface What is just outside the conductor ? - GE dl = 0 #. /11/11/11/1/8 S conductor E L > E. de + Bottom See +See t -un Exo 00 as d E. = 0 True for any such patha > - Ex = 0 > - Et surface outside conductor. S Gauss Law 1/ /// I /// ·Ed = Ed = +Gen = ~ O ELdAtEdA O as gro SEdE = dA True for any oa E = E= Poisson's Equation Recalling first Maxwell's equation : V. E(r) = gl · E(r) = & E(F) = - VV(r) -. -F(r) = I= Laplace's Equation ::v(r) If g(i) = 0 = 0 Uniqueness Theorem Suppose you have a closed surface' (enclosing VI. Suppose you are given the potential "Von , S. Then , there is only solution v(r) - one..E 1 : if ( , (r) and Ver) solve F = in V , and both agree with potential S then"v (r) = (r) given on , , NOTE : or its derivative E =V. n NOTE : equal to with in a constant if is given. Proof : Suppose( (r) V , , (r) are solutions ; let W = V. - Ve V = "V *V - = O ~ w - 7-7 divergence theorem St FdV Recalling :. =d With : ver) , f v(i) = v(r) = = + 1 (r)12 ↑ product rule Y (5) · XX() F soSivri)dV mosVi ~ = V and S VVVondA " = + - , Lan SIVvRdv ~ = 0 20 = > V= 0 everywhere in V W = constant Constant = 0 If V specified on S Since V = V at these points.  Sep 25 , 2024 Method of Images Uniqueness Theorem : Any solution that meets boundary conditions (on surface and has 5) same g(F) (inside S). Ex. az We want -(F) , E(r) for >O. HARD · q = plane at z = o 2 & o Negative O(X , Y) on surface L ↑ =G Z E(r) get a Considering W(r) and. +q easier in z 0 -V= O plane = d Surface : z= 0 plane and N done (closes at ). W & Both have v = 0 onS d Both have some g(r) ins vo-q -(r) =so (ir-co 11-i8-10 , , 0 , -all) Plugging in - (r) = 4+ ((x + yz + E - d(z( - (x + yz + (E + d(r)") E(F) = - VV(r) => A (XE p - my) 0 = EoEin w = E. Ez] z = o = Sox30((x2 yz + - + d d 2(31 - (x + y (x) E. n an surface d = (x2 + yz + d23 & Total charge on ground surface Soda = God cause Fire 4sdsd S= 0 = 25 (4 = - ) = -q & (sdyyO d Moving to a ground sphere closed volume betweenph and w I 9) z -(r) V (r · = , G, - & charges i I ~(r) = V(r , 0 4) = (v2 zi 2 rz , cost) "2 ori , + - so (rr 1 + z-2rzecose) 112 GN + Need -V /r = a , 0 , 4) = 0 X 0 , 9 & q + : 0 = COSO)12 (a + z-2azyCOSO (12 (a2 + z ? - 2az , a2 + z2-2azzcos = (E) (a + z-2az , c00) Need coefficient of cost equal : - 2az = -(2az = constant terms ar + zi = = (E)(an + zi) 0 I=F) =  Sep 27 , 2024 Laplace's Equation Poisson's Equation without & (charge density). (NO CHARGES) - V(F) 0 - = 2 dimensional : Vr) = (X Y) ,. No zaxis , so we set up that I is the same at all z. Recalling the Laplacian :2) =+ Solve subject to boundary conditions. Method of Separation of Variables Try f (x) fz(y) imm = + # = fz(y) f,(x) + f , (x)y4 +z(y) = 0 Divide by f , (x) fz(X) and more second term to right-hand side. f zu fz(y) - Indep. of Y Indep of X. > - Both constant e td f = k + z = -K defe = Kfn fe = Rfz Case 1 : k= 0 f, = (ax + b) fz = (cy + d) # (x , y) = (ax + b)(cy + d) = A + Bx + Cy +D Case 2 : k